Fundamentals of Compressible Fluid Mechanics

Genick Bar–Meir, Ph. D. 1107 16th Ave S. E. Minneapolis, MN 55414-2411 email:[email protected]

Copyright © 2006, 2005, and 2004 by Genick Bar-Meir See the ﬁle copying.fdl or copyright.tex for copying conditions. Version (0.4.4.2 aka 0.4.4.1j May 21, 2007) ‘We are like dwarfs sitting on the shoulders of giants”

from The Metalogicon by John in 1159 CONTENTS

GNU Free Documentation License ...... xvii 1. APPLICABILITY AND DEFINITIONS ...... xviii 2. VERBATIM COPYING ...... xix 3. COPYING IN QUANTITY ...... xix 4. MODIFICATIONS ...... xx 5. COMBINING DOCUMENTS ...... xxii 6. COLLECTIONS OF DOCUMENTS ...... xxii 7. AGGREGATION WITH INDEPENDENT WORKS ...... xxiii 8. TRANSLATION ...... xxiii 9. TERMINATION ...... xxiii 10. FUTURE REVISIONS OF THIS LICENSE ...... xxiii ADDENDUM: How to use this License for your documents ...... xxiv Potto Project License ...... xxv How to contribute to this book ...... xxvii Credits ...... xxvii John Martones ...... xxvii Grigory Toker ...... xxviii Ralph Menikoff ...... xxviii Your name here ...... xxviii Typo corrections and other ”minor” contributions ...... xxviii Version 0.4.3 Sep. 15, 2006 ...... xxxv Version 0.4.2 ...... xxxv Version 0.4 ...... xxxvi Version 0.3 ...... xxxvi Version 4.3 ...... xli Version 4.1.7 ...... xlii Speed of Sound ...... xlvi

iii iv CONTENTS Stagnation effects ...... xlvi Nozzle ...... xlvi Normal Shock ...... xlvi Isothermal Flow ...... xlvi Fanno Flow ...... xlvii Rayleigh Flow ...... xlvii Evacuation and ﬁlling semi rigid Chambers ...... xlvii Evacuating and ﬁlling chambers under external forces ...... xlvii Oblique Shock ...... xlvii Prandtl–Meyer ...... xlvii Transient problem ...... xlvii

1 Introduction 1 1.1 What is Compressible Flow ? ...... 1 1.2 Why Compressible Flow is Important? ...... 2 1.3 Historical Background ...... 2 1.3.1 Early Developments ...... 4 1.3.2 The shock wave puzzle ...... 5 1.3.3 Choking Flow ...... 9 1.3.4 External ﬂow ...... 13 1.3.5 Filling and Evacuating Gaseous Chambers ...... 15 1.3.6 Biographies of Major Figures ...... 15

2 Fundamentals of Basic Fluid Mechanics 25 2.1 Introduction ...... 25 2.2 Fluid Properties ...... 25 2.3 Control Volume ...... 25 2.4 Reynold’s Transport Theorem ...... 25

3 Speed of Sound 27 3.1 Motivation ...... 27 3.2 Introduction ...... 27 3.3 Speed of sound in ideal and perfect gases ...... 29 3.4 Speed of Sound in Real Gas ...... 31 3.5 Speed of Sound in Almost Incompressible Liquid ...... 35 3.6 Speed of Sound in Solids ...... 36 3.7 Sound Speed in Two Phase Medium ...... 37

4 Isentropic Flow 41 4.1 Stagnation State for Ideal Gas Model ...... 41 4.1.1 General Relationship ...... 41 4.1.2 Relationships for Small Mach Number ...... 44 4.2 Isentropic Converging-Diverging Flow in Cross Section ...... 45 4.2.1 The Properties in the Adiabatic Nozzle ...... 46 4.2.2 Isentropic Flow Examples ...... 50 CONTENTS v 4.2.3 Mass Flow Rate (Number) ...... 53 4.3 Isentropic Tables ...... 62 4.3.1 Isentropic Isothermal Flow Nozzle ...... 63 4.3.2 General Relationship ...... 63 4.4 The Impulse Function ...... 70 4.4.1 Impulse in Isentropic Adiabatic Nozzle ...... 70 4.4.2 The Impulse Function in Isothermal Nozzle ...... 73 4.5 Isothermal Table ...... 73 4.6 The effects of Real Gases ...... 74

5 Normal Shock 81 5.1 Solution of the Governing Equations ...... 84 5.1.1 Informal Model ...... 84 5.1.2 Formal Model ...... 84 5.1.3 Prandtl’s Condition ...... 88 5.2 Operating Equations and Analysis ...... 89 5.2.1 The Limitations of the Shock Wave ...... 90 5.2.2 Small Perturbation Solution ...... 90 5.2.3 Shock Thickness ...... 91 5.3 The Moving Shocks ...... 91 5.3.1 Shock Result from a Sudden and Complete Stop ...... 94 5.3.2 Moving Shock into Stationary Medium (Suddenly Open Valve) 96 5.3.3 Partially Open Valve ...... 101 5.3.4 Partially Closed Valve ...... 103 5.3.5 Worked–out Examples for Shock Dynamics ...... 104 5.4 Shock Tube ...... 109 5.5 Shock with Real Gases ...... 113 5.6 Shock in Wet Steam ...... 113 5.7 Normal Shock in Ducts ...... 113 5.8 More Examples for Moving Shocks ...... 114 5.9 Tables of Normal Shocks, k = 1.4 Ideal Gas ...... 115

6 Normal Shock in Variable Duct Areas 123 6.1 Nozzle efﬁciency ...... 129 6.2 Diffuser Efﬁciency ...... 129

7 Nozzle Flow With External Forces 135 7.1 Isentropic Nozzle (Q = 0) ...... 136 7.2 Isothermal Nozzle (T = constant) ...... 136

8 Isothermal Flow 137 8.1 The Control Volume Analysis/Governing equations ...... 138 8.2 Dimensionless Representation ...... 138 8.3 The Entrance Limitation of Supersonic Branch ...... 142 8.4 Comparison with Incompressible Flow ...... 143 vi CONTENTS 8.5 Supersonic Branch ...... 145 8.6 Figures and Tables ...... 146 8.7 Isothermal Flow Examples ...... 147 8.8 Unchoked situation ...... 152

9 Fanno Flow 155 9.1 Introduction ...... 155 9.2 Model ...... 156 9.3 Non–dimensionalization of the equations ...... 157 9.4 The Mechanics and Why the Flow is Choked? ...... 160 9.5 The working equations ...... 161 9.6 Examples of Fanno Flow ...... 164 9.7 Supersonic Branch ...... 169 9.8 Maximum length for the supersonic ﬂow ...... 169 9.9 Working Conditions ...... 170 4fL 9.9.1 Variations of The Tube Length ( D ) Effects ...... 171 9.9.2 The Pressure Ratio, P2 , effects ...... 176 P1 9.9.3 Entrance Mach number, M1, effects ...... 178 9.10 The Approximation of the Fanno ﬂow by Isothermal Flow ...... 185 9.11 More Examples of Fanno Flow ...... 186 9.12 The Table for Fanno Flow ...... 187

10 RAYLEIGH FLOW 189 10.1 Introduction ...... 189 10.2 Governing Equation ...... 190 10.3 Rayleigh Flow Tables ...... 193 10.4 Examples For Rayleigh Flow ...... 196

11 Evacuating and Filling a Semi Rigid Chambers 201 11.1 Governing Equations and Assumptions ...... 202 11.2 General Model and Non-dimensioned ...... 204 11.2.1 Isentropic Process ...... 205 11.2.2 Isothermal Process in The Chamber ...... 206 11.2.3 A Note on the Entrance Mach number ...... 206 11.3 Rigid Tank with Nozzle ...... 207 11.3.1 Adiabatic Isentropic Nozzle Attached ...... 207 11.3.2 Isothermal Nozzle Attached ...... 209 11.4 Rapid evacuating of a rigid tank ...... 209 11.4.1 With Fanno Flow ...... 209 11.4.2 Filling Process ...... 211 11.4.3 The Isothermal Process ...... 212 11.4.4 Simple Semi Rigid Chamber ...... 213 11.4.5 The “Simple” General Case ...... 213 11.5 Advance Topics ...... 215 CONTENTS vii

12 Evacuating/Filing Chambers under External Volume Control 217 12.1 General Model ...... 217 12.1.1 Rapid Process ...... 218 12.1.2 Examples ...... 221 12.1.3 Direct Connection ...... 221 12.2 Summary ...... 222

13 Topics in Unsteady one Dimensional gas dynamics 225

14 Oblique-Shock 227 14.1 Preface to Oblique Shock ...... 227 14.2 Introduction ...... 228 14.2.1 Introduction to Oblique Shock ...... 228 14.2.2 Introduction to Prandtl–Meyer Function ...... 228 14.2.3 Introduction to Zero Inclination ...... 229 14.3 Oblique Shock ...... 229 14.4 Solution of Mach Angle ...... 232 14.4.1 Upstream Mach Number, M1, and Deﬂection Angle, δ . . . . 232 14.4.2 When No Oblique Shock Exist or When D > 0 ...... 235 14.4.3 Upstream Mach Number, M1, and Shock Angle, θ ...... 241 14.4.4 Given Two Angles, δ and θ ...... 242 14.4.5 Flow in a Semi–2D Shape ...... 244 14.4.6 Small δ “Weak Oblique shock” ...... 244 14.4.7 Close and Far Views of the Oblique Shock ...... 245 14.4.8 Maximum Value of Oblique shock ...... 245 14.4.9 Detached shock ...... 246 14.4.10Issues Related to the Maximum Deﬂection Angle ...... 247 14.4.11Oblique Shock Examples ...... 249 14.4.12Application of Oblique Shock ...... 251 14.4.13Optimization of Suction Section Design ...... 262 14.5 Summary ...... 262 14.6 Appendix: Oblique Shock Stability Analysis ...... 263

15 Prandtl-Meyer Function 265 15.1 Introduction ...... 265 15.2 Geometrical Explanation ...... 266 15.2.1 Alternative Approach to Governing Equations ...... 267 15.2.2 Comparison And Limitations between the Two Approaches . 270 15.3 The Maximum Turning Angle ...... 271 15.4 The Working Equations for the Prandtl-Meyer Function ...... 271 15.5 d’Alembert’s Paradox ...... 272 15.6 Flat Body with an Angle of Attack ...... 273 15.7 Examples For Prandtl–Meyer Function ...... 274 15.8 Combination of the Oblique Shock and Isentropic Expansion . . . . 276 viii CONTENTS

16 Topics in Steady state Two Dimensional ﬂow 279

A Computer Program 281 A.1 About the Program ...... 281 A.2 Usage ...... 281 A.3 Program listings ...... 284

Index 285 Subjects index ...... 285 Authors index ...... 287 LIST OF FIGURES

1.1 The shock as connection of Fanno and Rayleigh lines after Stodola, Steam and Gas Turbine ...... 7 1.2 The schematic of deLavel’s turbine after Stodola, Steam and Gas Turbine ...... 9 1.3 The measured pressure in a nozzle taken from Stodola 1927 Steam and Gas Turbines ...... 11 1.4 Flow rate as a function of the back pressure taken from Stodola 1927 Steam and Gas Turbines ...... 12 1.5 Portrait of Galileo Galilei ...... 16 1.6 Photo of Ernest Mach ...... 17 1.7 The photo of thebullet in a supersonic ﬂow not taken in a wind tunnel 17 1.8 Photo of Lord Rayleigh ...... 18 1.9 Portrait of Rankine ...... 19 1.10 The photo of Gino Fanno approximately in 1950 ...... 20 1.11 Photo of Prandtl ...... 21 1.12 The photo of Ernst Rudolf George Eckert with the author’s family . . 22

3.1 A very slow moving piston in a still gas ...... 28 3.2 Stationary sound wave and gas moves relative to the pulse . . . . . 28 3.3 The Compressibility Chart ...... 32

4.1 Flow of a compressible substance (gas) through a converging–diverging nozzle...... 41 4.2 Perfect gas ﬂows through a tube ...... 43 4.3 The stagnation properties as a function of the Mach number, k = 1.4 44 4.4 Control volume inside a converging-diverging nozzle...... 46

ix x LIST OF FIGURES 4.5 The relationship between the cross section and the Mach number on the subsonic branch ...... 50 4.6 Various ratios as a function of Mach number for isothermal Nozzle . 66 4.7 The comparison of nozzle flow ...... 67 4.8 Comparison of the pressure and temperature drop as a function of the normalized length (two scales) ...... 68 4.9 Schematic to explain the significances of the Impulse function . . . . 71 4.10 Schematic of a flow of a compressible substance (gas) thorough a converging nozzle for example (4.7) ...... 72

5.1 A shock wave inside a tube, but it can also be viewed as a one– dimensional shock wave...... 81 5.2 The intersection of Fanno flow and Rayleigh flow produces two solutions for the shock wave...... 83 5.3 The exit Mach number and the stagnation pressure ratio as a function of upstream Mach number...... 87 5.4 The ratios of the static properties of the two sides of the shock. . . . 89 5.5 Comparison between stationary shock and moving shock in ducts . 91 5.6 Comparison between a stationary shock and a moving shock in a stationary medium in ducts...... 94 5.7 The moving shock Mach numbers as a result of a sudden and complete stop...... 95 5.8 A shock moves into a still medium as a result of a sudden and complete opening of a valve ...... 96 5.9 The number of iterations to achieve convergence...... 97 5.10 The maximum of “downstream” Mach number as a function of the specific heat, k...... 99 5.11 A shock moves into a moving medium as a result of a sudden and complete open valve...... 102 5.12 The results of the partial opening of the valve...... 103 5.13 A shock as a result of a sudden and partially a valve closing or a narrowing the passage to the flow ...... 103 5.14 Schematic of a piston pushing air in a tube...... 107 5.15 Figure for Example (5.8) ...... 109 5.16 The shock tube schematic with a pressure ”diagram.” ...... 110 5.17 Figure for Example (5.10) ...... 114 5.18 The results for Example (5.10) ...... 115

6.1 The flow in the nozzle with different back pressures ...... 123 6.2 A nozzle with normal shock ...... 124 6.3 Description to clarify the definition of diffuser efficiency ...... 130 6.4 Schematic of a supersonic tunnel in a continuous region (and also for example (6.3) ...... 130

8.1 Control volume for isothermal flow ...... 137 LIST OF FIGURES xi 8.2 Description of the pressure, temperature relationships as a function of the Mach number for isothermal flow ...... 143 8.3 The Mach number at the entrance to a tube under isothermal flow 4fL model as a function D ...... 153 9.1 Control volume of the gas flow in a constant cross section ...... 155 9.2 Various parameters in Fanno flow as a function of Mach number . . 163 9.3 Schematic of Example (9.1) ...... 164 9.4 The schematic of Example (9.2) ...... 166 9.5 The maximum length as a function of specific heat, k ...... 170 4fL 9.6 The effects of increase of D on the Fanno line ...... 171 9.7 The development properties in of converging nozzle ...... 172 9.8 The Mach numbers at entrance and exit of tube and mass flow rate 4fL for Fanno Flow as a function of the D ...... 173 4fL 9.9 M1 as a function M2 for various D ...... 174 4fL 9.10 M1 as a function M2 for different D for supersonic entrance velocity.175 4fL 4fL 9.11 The pressure distribution as a function of D for a short D . . . . 177 4fL 4fL 9.12 The pressure distribution as a function of D for a long D . . . . 178 9.13 The effects of pressure variations on Mach number profile as a func- 4fL 4fL tion of D when the total resistance D = 0.3 for Fanno Flow . . . 179 4fL 4fL 9.14 Fanno Flow Mach number as a function of D when the total D = 0.3 ...... 180 9.15 Schematic of a “long” tube in supersonic branch ...... 181 4fL 9.16 The extra tube length as a function of the shock location, D supersonic branch ...... 181 9.17 The maximum entrance Mach number, M1 to the tube as a function 4fL of D supersonic branch ...... 182 9.18 The entrance Mach number as a function of dimensionless resistance and comparison with Isothermal Flow ...... 185

10.1 The control volume of Rayleigh Flow ...... 189 10.2 The Temperature Entropy Diagram For Rayleigh Line ...... 191 10.3 The basic functions of Rayleigh Flow (k=1.4) ...... 195

11.1 The two different classifications of models that explain the filling or evacuating of a single chamber ...... 201 11.2 A schematic of two possible connections of the tube to a single chamber ...... 202 11.3 A schematic of the control volumes used in this model ...... 202 11.4 The pressure assumptions in the chamber and tube entrance . . . . 203 11.5 The reduced time as a function of the modified reduced pressure . . 210 11.6 The reduced time as a function of the modified reduced pressure . . 212

12.1 The control volume of the “Cylinder” ...... 218 xii LIST OF FIGURES 12.2 The pressure ratio as a function of the dimensionless time for chokeless condition ...... 223 12.3 The pressure ratio as a function of the dimensionless time for choked condition ...... 224 12.4 The pressure ratio as a function of the dimensionless time . . . . . 224

14.1 A view of a straight normal shock as a limited case for oblique shock 227 14.2 The regions where oblique shock or Prandtl–Meyer function exist. Notice that both have a maximum point and a “no solution” zone, which is around zero. However, Prandtl-Meyer function approaches closer to a zero deflection angle...... 228 14.3 A typical oblique shock schematic ...... 229 14.4 Flow around spherically blunted 30◦ cone-cylinder with Mach number 2.0. It can be noticed that the normal shock, the strong shock, and the weak shock coexist...... 235 14.5 The view of a large inclination angle from different points in the fluid field...... 236 14.6 The various coefficients of three different Mach numbers to demonstrate that D is zero ...... 239 14.7 The Mach waves that are supposed to be generated at zero inclination.240 14.8 The calculation of D (possible error), shock angle, and exit Mach number for M1 = 3 ...... 241 14.9 The possible range of solutions for different parameters for given upstream Mach numbers ...... 243 14.10Schematic of finite wedge with zero angle of attack...... 244 14.11A local and a far view of the oblique shock...... 245 14.12The schematic for a round–tip bullet in a supersonic flow ...... 247 14.13The schematic for a symmetrical suction section with Mach reflection 248 14.14The “detached” shock in a complicated configuration sometimes referred to as Mach reflection ...... 248 14.15Oblique shock occurs around a cone. This photo is courtesy of Dr. Grigory Toker, a Research Professor at Cuernavaco University of Mexico. According to his measurement, the cone half angle is 15◦ and the Mach number is 2.2...... 250 14.16Maximum values of the properties in an oblique shock ...... 251 14.17Two variations of inlet suction for supersonic flow ...... 251 14.18Schematic for Example (14.4) ...... 252 14.19Schematic for Example (14.5) ...... 253 14.20Schematic of two angles turn with two weak shocks ...... 254 14.21Typical examples of unstable and stable situations ...... 263 14.22The schematic of stability analysis for oblique shock...... 264

15.1 The definition of the angle for the Prandtl–Meyer function...... 265 15.2 The angles of the Mach line triangle ...... 265 15.3 The schematic of the turning flow ...... 266 LIST OF FIGURES xiii 15.4 The schematic of the coordinate based on the mathematical description ...... 267 15.5 Expansion of Prandtl-Meyer function when it exceeds the maximum angle ...... 272 15.7 A simplified diamond shape to illustrate the supersonic d’Alembert’s Paradox ...... 272 15.6 The angle as a function of the Mach number ...... 273 15.8 The definition of the angle for the Prandtl–Meyer function...... 273 15.9 The schematic of Example 15.1 ...... 274 15.10The schematic for the reversed question of example (15.2) . . . . . 275

A.1 Schematic diagram that explains the structure of the program . . . . 282 xiv LIST OF FIGURES LIST OF TABLES

3.1 Water speed of sound from different sources ...... 35 3.2 Liquids speed of sound, after Aldred, John, Manual of Sound Record- ing, London: Fountain Press, 1972 ...... 36 3.3 Solids speed of sound, after Aldred, John, Manual of Sound Record- ing, London:Fountain Press, 1972 ...... 37

4.1 Fliegner’s number and other paramters as function of Mach number 58 4.1 Fliegner’s number and other paramters as function of Mach number (continue) ...... 59 4.1 Fliegner’s number and other paramters as function of Mach number (continue) ...... 60 4.2 Isentropic Table k = 1.4 ...... 62 4.2 Isentropic Table k=1.4 (continue) ...... 63 4.3 Isothermal Table ...... 73 4.3 Isothermal Table (continue) ...... 74

5.1 The shock wave table for k = 1.4 ...... 115 5.1 The shock wave table for k = 1.4 (continue) ...... 116 5.1 The shock wave table for k = 1.4 (continue) ...... 117 5.2 Table for a Reﬂective Shock from a suddenly closed end (k=1.4) . . 117 5.2 Table for Reﬂective Shock from suddenly closed valve (end) (k=1.4)(continue)118 5.3 Table for shock propagating from suddenly opened valve (k=1.4) . . 118 5.3 Table for shock propagating from suddenly opened valve (k=1.4) . . 119 5.4 Table for shock propagating from a suddenly opened valve (k=1.3) 119 5.4 Table for shock propagating from a suddenly opened valve (k=1.3) 120 5.4 Table for shock propagating from a suddenly opened valve (k=1.3) 121

xv xvi LIST OF TABLES 8.1 The Isothermal Flow basic parameters ...... 147 8.4 The ﬂow parameters for unchoked ﬂow ...... 152

9.1 Fanno Flow Standard basic Table ...... 187 9.1 Fanno Flow Standard basic Table (continue) ...... 188

10.1 Rayleigh Flow k=1.4 ...... 193 10.1 Rayleigh Flow k=1.4 (continue) ...... 194 10.1 Rayleigh Flow k=1.4 (continue) ...... 195

14.1 Table of maximum values of the oblique Shock k=1.4 ...... 245 14.1 Maximum values of oblique shock (continue) k=1.4 ...... 246 Notice of Copyright For This Document:

This document is published under dual licenses: You can choose the license under which you use the document and associate ﬁles and software.

51 Franklin St, Fifth Floor, Boston, MA 02110-1301 USA

Everyone is permitted to copy and distribute verbatim copies of this license document, but changing it is not allowed. Preamble The purpose of this License is to make a manual, textbook, or other functional and useful document ”free” in the sense of freedom: to assure everyone the effective freedom to copy and redistribute it, with or without modifying it, either commercially or noncommercially. Secondarily, this License preserves for the author and publisher a way to get credit for their work, while not being considered responsible for modiﬁcations made by others. This License is a kind of ”copyleft”, which means that derivative works of the document must themselves be free in the same sense. It complements the GNU General Public License, which is a copyleft license designed for free software. We have designed this License in order to use it for manuals for free software, because free software needs free documentation: a free program should

xvii xviii LIST OF TABLES come with manuals providing the same freedoms that the software does. But this License is not limited to software manuals; it can be used for any textual work, regardless of subject matter or whether it is published as a printed book. We recommend this License principally for works whose purpose is instruction or reference.

1. APPLICABILITY AND DEFINITIONS

This License applies to any manual or other work, in any medium, that contains a notice placed by the copyright holder saying it can be distributed under the terms of this License. Such a notice grants a world-wide, royalty-free license, unlimited in duration, to use that work under the conditions stated herein. The ”Document”, below, refers to any such manual or work. Any member of the public is a licensee, and is addressed as ”you”. You accept the license if you copy, modify or distribute the work in a way requiring permission under copyright law. A ”Modified Version” of the Document means any work containing the Document or a portion of it, either copied verbatim, or with modifications and/or translated into another language. A ”Secondary Section” is a named appendix or a front-matter section of the Document that deals exclusively with the relationship of the publishers or authors of the Document to the Document’s overall subject (or to related matters) and contains nothing that could fall directly within that overall subject. (Thus, if the Document is in part a textbook of mathematics, a Secondary Section may not explain any mathematics.) The relationship could be a matter of historical connection with the subject or with related matters, or of legal, commercial, philosophical, ethical or political position regarding them. The ”Invariant Sections” are certain Secondary Sections whose titles are designated, as being those of Invariant Sections, in the notice that says that the Document is released under this License. If a section does not fit the above definition of Secondary then it is not allowed to be designated as Invariant. The Document may contain zero Invariant Sections. If the Document does not identify any Invariant Sections then there are none. The ”Cover Texts” are certain short passages of text that are listed, as Front-Cover Texts or Back-Cover Texts, in the notice that says that the Document is released under this License. A Front-Cover Text may be at most 5 words, and a Back-Cover Text may be at most 25 words. A ”Transparent” copy of the Document means a machine-readable copy, represented in a format whose specification is available to the general public, that is suitable for revising the document straightforwardly with generic text editors or (for images composed of pixels) generic paint programs or (for drawings) some widely available drawing editor, and that is suitable for input to text formatters or for automatic translation to a variety of formats suitable for input to text formatters. A copy made in an otherwise Transparent file format whose markup, or absence of markup, has been arranged to thwart or discourage subsequent modification GNU FREE DOCUMENTATION LICENSE xix by readers is not Transparent. An image format is not Transparent if used for any substantial amount of text. A copy that is not ”Transparent” is called ”Opaque”. Examples of suitable formats for Transparent copies include plain ASCII without markup, Texinfo input format, LaTeX input format, SGML or XML using a publicly available DTD, and standard-conforming simple HTML, PostScript or PDF designed for human modification. Examples of transparent image formats include PNG, XCF and JPG. Opaque formats include proprietary formats that can be read and edited only by proprietary word processors, SGML or XML for which the DTD and/or processing tools are not generally available, and the machine-generated HTML, PostScript or PDF produced by some word processors for output purposes only. The ”Title Page” means, for a printed book, the title page itself, plus such following pages as are needed to hold, legibly, the material this License requires to appear in the title page. For works in formats which do not have any title page as such, ”Title Page” means the text near the most prominent appearance of the work’s title, preceding the beginning of the body of the text. A section ”Entitled XYZ” means a named subunit of the Document whose title either is precisely XYZ or contains XYZ in parentheses following text that translates XYZ in another language. (Here XYZ stands for a specific section name mentioned below, such as ”Acknowledgements”, ”Dedications”, ”En- dorsements”, or ”History”.) To ”Preserve the Title” of such a section when you modify the Document means that it remains a section ”Entitled XYZ” according to this definition. The Document may include Warranty Disclaimers next to the notice which states that this License applies to the Document. These Warranty Disclaimers are considered to be included by reference in this License, but only as regards dis- claiming warranties: any other implication that these Warranty Disclaimers may have is void and has no effect on the meaning of this License. 2. VERBATIM COPYING You may copy and distribute the Document in any medium, either commercially or noncommercially, provided that this License, the copyright notices, and the license notice saying this License applies to the Document are reproduced in all copies, and that you add no other conditions whatsoever to those of this Li- cense. You may not use technical measures to obstruct or control the reading or further copying of the copies you make or distribute. However, you may accept compensation in exchange for copies. If you distribute a large enough number of copies you must also follow the conditions in section 3. You may also lend copies, under the same conditions stated above, and you may publicly display copies. 3. COPYING IN QUANTITY If you publish printed copies (or copies in media that commonly have printed covers) of the Document, numbering more than 100, and the Document’s xx LIST OF TABLES license notice requires Cover Texts, you must enclose the copies in covers that carry, clearly and legibly, all these Cover Texts: Front-Cover Texts on the front cover, and Back-Cover Texts on the back cover. Both covers must also clearly and legibly identify you as the publisher of these copies. The front cover must present the full title with all words of the title equally prominent and visible. You may add other material on the covers in addition. Copying with changes limited to the covers, as long as they preserve the title of the Document and satisfy these conditions, can be treated as verbatim copying in other respects. If the required texts for either cover are too voluminous to fit legibly, you should put the first ones listed (as many as fit reasonably) on the actual cover, and continue the rest onto adjacent pages. If you publish or distribute Opaque copies of the Document numbering more than 100, you must either include a machine-readable Transparent copy along with each Opaque copy, or state in or with each Opaque copy a computer- network location from which the general network-using public has access to download using public-standard network protocols a complete Transparent copy of the Document, free of added material. If you use the latter option, you must take reasonably prudent steps, when you begin distribution of Opaque copies in quantity, to ensure that this Transparent copy will remain thus accessible at the stated location until at least one year after the last time you distribute an Opaque copy (directly or through your agents or retailers) of that edition to the public. It is requested, but not required, that you contact the authors of the Doc- ument well before redistributing any large number of copies, to give them a chance to provide you with an updated version of the Document. 4. MODIFICATIONS You may copy and distribute a Modified Version of the Document under the conditions of sections 2 and 3 above, provided that you release the Modified Version under precisely this License, with the Modified Version filling the role of the Document, thus licensing distribution and modification of the Modified Version to whoever possesses a copy of it. In addition, you must do these things in the Modified Version: A. Use in the Title Page (and on the covers, if any) a title distinct from that of the Document, and from those of previous versions (which should, if there were any, be listed in the History section of the Document). You may use the same title as a previous version if the original publisher of that version gives permission. B. List on the Title Page, as authors, one or more persons or entities responsible for authorship of the modifications in the Modified Version, together with at least five of the principal authors of the Document (all of its principal authors, if it has fewer than five), unless they release you from this requirement. C. State on the Title page the name of the publisher of the Modified Version, as the publisher. GNU FREE DOCUMENTATION LICENSE xxi D. Preserve all the copyright notices of the Document. E. Add an appropriate copyright notice for your modifications adjacent to the other copyright notices. F. Include, immediately after the copyright notices, a license notice giving the public permission to use the Modified Version under the terms of this License, in the form shown in the Addendum below. G. Preserve in that license notice the full lists of Invariant Sections and required Cover Texts given in the Document’s license notice. H. Include an unaltered copy of this License. I. Preserve the section Entitled ”History”, Preserve its Title, and add to it an item stating at least the title, year, new authors, and publisher of the Modified Version as given on the Title Page. If there is no section Entitled ”History” in the Document, create one stating the title, year, authors, and publisher of the Document as given on its Title Page, then add an item describing the Modified Version as stated in the previous sentence. J. Preserve the network location, if any, given in the Document for public access to a Transparent copy of the Document, and likewise the network locations given in the Document for previous versions it was based on. These may be placed in the ”History” section. You may omit a network location for a work that was published at least four years before the Document itself, or if the original publisher of the version it refers to gives permission. K. For any section Entitled ”Acknowledgements” or ”Dedications”, Preserve the Title of the section, and preserve in the section all the substance and tone of each of the contributor acknowledgements and/or dedications given therein. L. Preserve all the Invariant Sections of the Document, unaltered in their text and in their titles. Section numbers or the equivalent are not considered part of the section titles. M. Delete any section Entitled ”Endorsements”. Such a section may not be included in the Modified Version. N. Do not retitle any existing section to be Entitled ”Endorsements” or to conflict in title with any Invariant Section. O. Preserve any Warranty Disclaimers.

If the Modified Version includes new front-matter sections or appendices that qualify as Secondary Sections and contain no material copied from the Docu- ment, you may at your option designate some or all of these sections as invariant. To do this, add their titles to the list of Invariant Sections in the Modified Version’s license notice. These titles must be distinct from any other section titles. xxii LIST OF TABLES You may add a section Entitled ”Endorsements”, provided it contains nothing but endorsements of your Modified Version by various parties–for example, statements of peer review or that the text has been approved by an organization as the authoritative definition of a standard. You may add a passage of up to five words as a Front-Cover Text, and a passage of up to 25 words as a Back-Cover Text, to the end of the list of Cover Texts in the Modified Version. Only one passage of Front-Cover Text and one of Back- Cover Text may be added by (or through arrangements made by) any one entity. If the Document already includes a cover text for the same cover, previously added by you or by arrangement made by the same entity you are acting on behalf of, you may not add another; but you may replace the old one, on explicit permission from the previous publisher that added the old one. The author(s) and publisher(s) of the Document do not by this License give permission to use their names for publicity for or to assert or imply endorsement of any Modified Version. 5. COMBINING DOCUMENTS

You may combine the Document with other documents released under this License, under the terms defined in section 4 above for modified versions, provided that you include in the combination all of the Invariant Sections of all of the original documents, unmodified, and list them all as Invariant Sections of your combined work in its license notice, and that you preserve all their Warranty Disclaimers. The combined work need only contain one copy of this License, and multiple identical Invariant Sections may be replaced with a single copy. If there are multiple Invariant Sections with the same name but different contents, make the title of each such section unique by adding at the end of it, in parentheses, the name of the original author or publisher of that section if known, or else a unique number. Make the same adjustment to the section titles in the list of Invariant Sections in the license notice of the combined work. In the combination, you must combine any sections Entitled ”History” in the various original documents, forming one section Entitled ”History”; likewise combine any sections Entitled ”Acknowledgements”, and any sections Entitled ”Dedications”. You must delete all sections Entitled ”Endorsements”. 6. COLLECTIONS OF DOCUMENTS

You may make a collection consisting of the Document and other documents released under this License, and replace the individual copies of this Li- cense in the various documents with a single copy that is included in the collection, provided that you follow the rules of this License for verbatim copying of each of the documents in all other respects. You may extract a single document from such a collection, and distribute it individually under this License, provided you insert a copy of this License into GNU FREE DOCUMENTATION LICENSE xxiii the extracted document, and follow this License in all other respects regarding verbatim copying of that document.

7. AGGREGATION WITH INDEPENDENT WORKS

A compilation of the Document or its derivatives with other separate and independent documents or works, in or on a volume of a storage or distribution medium, is called an ”aggregate” if the copyright resulting from the compilation is not used to limit the legal rights of the compilation’s users beyond what the individual works permit. When the Document is included in an aggregate, this Li- cense does not apply to the other works in the aggregate which are not themselves derivative works of the Document. If the Cover Text requirement of section 3 is applicable to these copies of the Document, then if the Document is less than one half of the entire aggregate, the Document’s Cover Texts may be placed on covers that bracket the Document within the aggregate, or the electronic equivalent of covers if the Document is in electronic form. Otherwise they must appear on printed covers that bracket the whole aggregate.

8. TRANSLATION

Translation is considered a kind of modiﬁcation, so you may distribute translations of the Document under the terms of section 4. Replacing Invariant Sections with translations requires special permission from their copyright holders, but you may include translations of some or all Invariant Sections in addition to the original versions of these Invariant Sections. You may include a translation of this License, and all the license notices in the Document, and any Warranty Dis- claimers, provided that you also include the original English version of this License and the original versions of those notices and disclaimers. In case of a disagree- ment between the translation and the original version of this License or a notice or disclaimer, the original version will prevail. If a section in the Document is Entitled ”Acknowledgements”, ”Dedica- tions”, or ”History”, the requirement (section 4) to Preserve its Title (section 1) will typically require changing the actual title.

9. TERMINATION

You may not copy, modify, sublicense, or distribute the Document except as expressly provided for under this License. Any other attempt to copy, modify, sublicense or distribute the Document is void, and will automatically terminate your rights under this License. However, parties who have received copies, or rights, from you under this License will not have their licenses terminated so long as such parties remain in full compliance.

10. FUTURE REVISIONS OF THIS LICENSE xxiv LIST OF TABLES The Free Software Foundation may publish new, revised versions of the GNU Free Documentation License from time to time. Such new versions will be similar in spirit to the present version, but may differ in detail to address new problems or concerns. See http://www.gnu.org/copyleft/. Each version of the License is given a distinguishing version number. If the Document speciﬁes that a particular numbered version of this License ”or any later version” applies to it, you have the option of following the terms and conditions either of that speciﬁed version or of any later version that has been published (not as a draft) by the Free Software Foundation. If the Document does not specify a version number of this License, you may choose any version ever published (not as a draft) by the Free Software Foundation.

ADDENDUM: How to use this License for your documents

To use this License in a document you have written, include a copy of the License in the document and put the following copyright and license notices just after the title page:

Copyright ©YEAR YOUR NAME. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation; with no Invariant Sections, no Front- Cover Texts, and no Back-Cover Texts. A copy of the license is included in the section entitled ”GNU Free Documentation License”.

If you have Invariant Sections, Front-Cover Texts and Back-Cover Texts, replace the ”with...Texts.” line with this:

with the Invariant Sections being LIST THEIR TITLES, with the Front- Cover Texts being LIST, and with the Back-Cover Texts being LIST.

If you have Invariant Sections without Cover Texts, or some other combination of the three, merge those two alternatives to suit the situation. If your document contains nontrivial examples of program code, we recommend releasing these examples in parallel under your choice of free software license, such as the GNU General Public License, to permit their use in free software. POTTO PROJECT LICENSE xxv Potto Project License This document may be redistributed provided a pointer appears in a prominent place showing clearly where the original version was published and/or was obtained. The original version of this document may be found at http://www. potto.org/copyright.html This document is derived from open content license http://opencontent. org/opl.shtml LICENSE Terms and Conditions for Copying, Distributing, and Modifying

1. Disclaimer of warranty of the original author You may copy and distribute exact replicas of this document as you receive it, in any medium, provided that you conspicuously and appropriately publish on each copy an appropriate copyright notice and disclaimer of warranty of the original author; keep intact all the copyright notices that refer to this document. You may at your discretion charge a fee for the media and/or handling involved in creating a unique copy of this document. You may offer instruc- tional support for this document and software exchange for a fee. You may at your option offer warranty in exchange for a fee.

2. Modification and distribution of modified material You may modify your copy or copies of this document and the attached software or any portion of it. You may distribute such modifications, all the material based on this original content or work, under the terms of Section 1 above.

3. Your Name and Communication With You If you wish to modify this text or software in any way, you must document the nature of those modiﬁcations in the ”Credits” section along with your name, and information concerning how you may be contacted. You must have a reasonable way to contact you.

4. No Endorsement The names ”POTTO Project” and ”Fundamentals of Compressible Fluid Me- chanics” or the author of this document must not be used to endorse or promote products derived from this text (book or software) without prior written permission.

5. Derived Name(s) Products derived from this software may not be called “POTTO Project,” or alleged association with this author nor may “POTTO” or “POTTO Project” appear in their name, without prior written permission of the Dr. Genick Bar- Meir. xxvi LIST OF TABLES 6. Applicability of this license You are not required to accept this License, since you have not signed it. However, nothing else grants you permission to copy, distribute or modify these materials. These actions are prohibited by law if you do not accept this License. Therefore, by distributing or translating these materials, or by deriving works here from, you indicate your acceptance of this License to do so, and all its terms and conditions for copying, distributing or translating these materials. 7. No Warranty Because these materials are licensed free of charge, there is no warranty for the manuscript, to the extent permitted by applicable law. Except when otherwise stated in writing the copyright holders and/or other parties provide these manuscripts “AS IS” without warranty of any kind, either expressed or implied, including, but not limited to, the implied warranties of merchantability and ﬁtness for a particular purpose. The entire risk of use of this manuscript is with you. Should this manuscript prove faulty, inaccurate, or otherwise unacceptable you assume the cost of all necessary repair or correction. 8. No Liability In no event unless required by applicable law or agreed to in writing will any copyright holder, or any other party who may mirror and/or redistribute these materials as permitted above, be liable to you for damages, including any general, special, incidental or consequential damages arising out of the use or inability to use this manuscript, even if such holder or other party has been advised of the possibility of such damages. 9. Jurisdiction These terms and conditions are governed by and will be interpreted in ac- cordance with the state of POTTO Project residence law and any disputes relating to these terms and conditions will be exclusively subject to the jurisdiction of the courts of POTTO Project residence. Currently, the POTTO Project residence is the state of Minnesota. The various provisions of these terms and conditions are severable and if any provision is held to be invalid or unenforceable by any court of competent jurisdiction then such invalidity or unenforceability shall not affect the remaining provisions. If these terms and conditions are not accepted in full, you use the book and or the software must be terminated immediately. CONTRIBUTOR LIST

How to contribute to this book As a copylefted work, this book is open to revision and expansion by any interested parties. The only ”catch” is that credit must be given where credit is due. This is a copyrighted work: it is not in the public domain! If you wish to cite portions of this book in a work of your own, you must follow the same guidelines as for any other GDL copyrighted work.

Credits All entries arranged in alphabetical order of surname. Major contributions are listed by individual name with some detail on the nature of the contribution(s), date, contact info, etc. Minor contributions (typo corrections, etc.) are listed by name only for reasons of brevity. Please understand that when I classify a contribution as ”minor,” it is in no way inferior to the effort or value of a ”major” contribution, just smaller in the sense of less text changed. Any and all contributions are gratefully accepted. I am indebted to all those who have given freely of their own knowledge, time, and resources to make this a better book!

• Date(s) of contribution(s): 2004 to present

• Nature of contribution: Original author.

• Contact at: [email protected]

John Martones • Date(s) of contribution(s): June 2005

xxvii xxviii LIST OF TABLES

• Nature of contribution: HTML formatting, some error corrections.

Grigory Toker • Date(s) of contribution(s): August 2005 • Nature of contribution: Provided pictures of the oblique shock for oblique shcok chapter.

Ralph Menikoff • Date(s) of contribution(s): July 2005 • Nature of contribution: Some discussion about the solution to oblique shock and about the Maximum Deﬂection of the oblique shock.

Your name here • Date(s) of contribution(s): Month and year of contribution • Nature of contribution: Insert text here, describing how you contributed to the book.

• Contact at: my [email protected]

Typo corrections and other ”minor” contributions • H. Gohrah, Ph. D., September 2005, some LaTeX issues. • Roy Tate November 2006, Suggestions on improving english and gramer. About This Author

Genick Bar-Meir holds a Ph.D. in Mechanical Engineering from University of Min- nesota and a Master in Fluid Mechanics from Tel Aviv University. Dr. Bar-Meir was the last student of the late Dr. R.G.E. Eckert. Much of his time has been spend doing research in the field of heat and mass transfer (this includes fluid mechanics) related to manufacturing processes and design. Currently, he spends time writing books and software for the POTTO project (see Potto Prologue). The author enjoys to encourages his students to understand the material beyond the basic requirements of exams. In his early part of his professional life, Bar-Meir was mainly interested in elegant models whether they have or not a practical applicability. Now, this author’s views had changed and the virtue of the practical part of any model becomes the essential part of his ideas, books and softwares. He developed models for Mass Transfer in high concentration that became a building blocks for many other models. These models are based on analytical solution to a family of equations1. As the change in the view occurred, Bar-Meir developed models that explained several manufacturing processes such the rapid evacuation of gas from containers, the critical piston velocity in a partially filled chamber (related to hydraulic jump), supply and demand to rapid change power system and etc. All the models have practical applicability. These models have been extended by several research groups (need- less to say with large research grants). For example, the Spanish Comision Inter- ministerial provides grants TAP97-0489 and PB98-0007, and the CICYT and the European Commission provides 1FD97-2333 grants for minor aspects of that models. Moreover, the author’s models were used in numerical works, in GM, British industry, Spain, and even Iran. The author believes that this book, as in the past, will promote new re-

1Where the mathematicians were able only to prove that the solution exists.

xxix xxx LIST OF TABLES search. More than that, this author believes that the book will blaze a trail of new understanding. The author lives with his wife and three children. A past project of his was building a four stories house, practically from scratch. While he writes his programs and does other computer chores, he often feels clueless about computers and programing. While he known to look like he know about many things, the author just know to learn quickly. The author spent years working on the sea (ships) as a engine sea ofﬁcer but now the author prefers to remain on solid ground. Prologue For The POTTO Project

This series of books was born out of frustrations in two respects. The first issue is the enormous price of college textbooks. It is unacceptable that the price of the college books will be over $150 per book (over 10 hours of work for an average student in The United States). The second issue that prompted the writing of this book is the fact that we as the public have to deal with a corrupted judicial system. As individuals we have to obey the law, particularly the copyright law with the “infinite2” time with the copyright holders. However, when applied to “small” individuals who are not able to hire a large legal firm, judges simply manufacture facts to make the little guy lose and pay for the defense of his work. On one hand, the corrupted court system defends the “big” guys and on the other hand, punishes the small “entrepreneur” who tries to defend his or her work. It has become very clear to the author and founder of the POTTO Project that this situation must be stopped. Hence, the creation of the POTTO Project. As R. Kook, one of this author’s sages, said instead of whining about arrogance and incorrectness, one should increase wisdom. This project is to increase wisdom and humility. The POTTO Project has far greater goals than simply correcting an abusive Judicial system or simply exposing abusive judges. It is apparent that writing textbooks especially for college students as a cooperation, like an open source, is a new idea3. Writing a book in the technical field is not the same as writing a novel. The writing of a technical book is really a collection of information and practice. There is always someone who can add to the book. The study of technical

2After the last decision of the Supreme Court in the case of Eldred v. Ashcroff (see http://cyber. law.harvard.edu/openlaw/eldredvashcroft for more information) copyrights practically remain in- deﬁnitely with the holder (not the creator). 3In some sense one can view the encyclopedia Wikipedia as an open content project (see http: //en.wikipedia.org/wiki/Main Page). The wikipedia is an excellent collection of articles which are written by various individuals.

xxxi xxxii LIST OF TABLES material isn’t only done by having to memorize the material, but also by coming to understand and be able to solve related problems. The author has not found any technique that is more useful for this purpose than practicing the solving of problems and exercises. One can be successful when one solves as many problems as possible. To reach this possibility the collective book idea was created/adapted. While one can be as creative as possible, there are always others who can see new aspects of or add to the material. The collective material is much richer than any single person can create by himself. The following example explains this point: The army ant is a kind of carnivorous ant that lives and hunts in the tropics, hunting animals that are even up to a hundred kilograms in weight. The secret of the ants’ power lies in their collective intelligence. While a single ant is not intelligent enough to attack and hunt large prey, the collective power of their networking creates an extremely powerful intelligence to carry out this attack ( see for information http://www.ex.ac.uk/ bugclub/raiders.html)4. So when an insect which is blind can be so powerful by networking, so can we in creating textbooks by this powerful tool. Why would someone volunteer to be an author or organizer of such a book? This is the first question the undersigned was asked. The answer varies from individual to individual. It is hoped that because of the open nature of these books, they will become the most popular books and the most read books in their respected field. In a way, the popularity of the books should be one of the incentives for potential contributors. The desire to be an author of a well-known book (at least in his/her profession) will convince some to put forth the effort. For some authors, the reason is the pure fun of writing and organizing educational material. Experience has shown that in explaining to others any given subject, one also begins to better understand the material. Thus, contributing to this book will help one to understand the material better. For others, the writing of or contributing to this kind of book will serve as a social function. The social function can have at least two components. One component is to come to know and socialize with many in the profession. For others the social part is as simple as a desire to reduce the price of college textbooks, especially for family members or relatives and those students lacking funds. For some contributors/authors, in the course of their teaching they have found that the textbook they were using contains sections that can be improved or that are not as good as their own notes. In these cases, they now have an opportunity to put their notes to use for others. Whatever the reasons, the undersigned believes that personal intentions are appropriate and are the author’s/organizer’s private affair. If a contributor of a section in such a book can be easily identified, then that contributor will be the copyright holder of that specific section (even within question/answer sections). The book’s contributor’s names could be written by their sections. It is not just for experts to contribute, but also students who happened to be doing their homework. The student’s contributions can be done by

4see also in Franks, Nigel R.; ”Army Ants: A Collective Intelligence,” American Scientist, 77:139, 1989 CREDITS xxxiii adding a question and perhaps the solution. Thus, this method is expected to accelerate the creation of these high quality books. These books are written in a similar manner to the open source software process. Someone has to write the skeleton and hopefully others will add “flesh and skin.” In this process, chapters or sections can be added after the skeleton has been written. It is also hoped that others will contribute to the question and answer sections in the book. But more than that, other books contain data5 which can be typeset in LATEX. These data (tables, graphs and etc.) can be redone by anyone who has the time to do it. Thus, the contributions to books can be done by many who are not experts. Additionally, contributions can be made from any part of the world by those who wish to translate the book. It is hoped that the book will be error-free. Nevertheless, some errors are possible and expected. Even if not complete, better discussions or better explanations are all welcome to these books. These books are intended to be “continuous” in the sense that there will be someone who will maintain and improve the book with time (the organizer). These books should be considered more as a project than to fit the traditional definition of “plain” books. Thus, the traditional role of author will be replaced by an organizer who will be the one to compile the book. The organizer of the book in some instances will be the main author of the work, while in other cases This may merely be the person who decides what will go into the book and what will not (gate keeper). Unlike a regular book, these works will have a version number because they are alive and continuously evolving. The undersigned of this document intends to be the organizer/author/coordinator of the projects in the following areas:

project name progress remarks version Die Casting alpha 0.0.3 Mechanics not started yet 0.0.0 Statics not started yet 0.0.0 Dynamics not started yet 0.0.0 Strength of Material not started yet 0.0.0 Compressible Flow early beta 0.4 Fluid Mechanics alpha 0.1 Thermodynamics early alpha 0.0.01 Heat Transfer not started yet Based on Eckert 0.0.0 Open Channel Flow not started yet 0.0.0 Two/Multi phases ﬂow not started yet Tel-Aviv’notes 0.0.0

The meaning of the progress is as:

• The Alpha Stage is when some of the chapters are already in rough draft;

5Data are not copyrighted. xxxiv LIST OF TABLES • In Beta Stage is when all or almost all of the chapters have been written and are at least in a draft stage; and

• In Gamma Stage is when all the chapters are written and some of the chapters are in a mature form. • The Advanced Stage is when all of the basic material is written and all that is left are aspects that are active, advanced topics, and special cases. The mature stage of a chapter is when all or nearly all of the sections are in a mature stage and have a mature bibliography as well as mature and numerous examples for every section. The mature stage of a section is when all of the topics in the section are written, and all of the examples and data (tables, figures, etc.) are already presented. While some terms are defined in a relatively clear fashion, other definitions give merely a hint on the status. But such a thing is hard to define and should be enough for this stage. The idea that a book can be created as a project has mushroomed from the open source software concept, but it has roots in the way science progresses. However, traditionally books have been improved by the same author(s), a process in which books have a new version every a few years. There are book(s) that have continued after their author passed away, i.e., the Boundary Layer Theory originated6 by Hermann Schlichting but continues to this day. However, projects such as the Linux Documentation project demonstrated that books can be written as the cooperative effort of many individuals, many of whom volunteered to help. Writing a textbook is comprised of many aspects, which include the actual writing of the text, writing examples, creating diagrams and figures, and writing 7 the LATEX macros which will put the text into an attractive format. These chores can be done independently from each other and by more than one individual. Again, because of the open nature of this project, pieces of material and data can be used by different books.

6Originally authored by Dr. Schlichting, who passed way some years ago. A new version is created every several years. 7One can only expect that open source and readable format will be used for this project. But more than that, only LATEX, and perhaps troff, have the ability to produce the quality that one expects for these writings. The text processes, especially LATEX, are the only ones which have a cross platform ability to produce macros and a uniform feel and quality. Word processors, such as OpenOfﬁce, Abiword, and Microsoft Word software, are not appropriate for these projects. Further, any text that is produced by Microsoft and kept in “Microsoft” format are against the spirit of this project In that they force spending money on Microsoft software. Prologue For This Book

Version 0.4.3 Sep. 15, 2006 The title of this section is change to reflect that it moved to beginning of the book. While it moves earlier but the name was not changed. Dr. Menikoff pointed to this inconsistency, and the author is apologizing for this omission. Several sections were add to this book with many new ideas for example on the moving shock tables. However, this author cannot add all the things that he was asked and want to the book in instant fashion. For example, one of the reader ask why not one of the example of oblique shock was not turn into the explanation of von Neumann paradox. The author was asked by a former client why he didn’t insert his improved tank filling and evacuating models (the addtion of the energy equation instead of isentropic model). While all these requests are important, the time is limited and they will be inserted as time permitted. The moving shock issues are not completed and more work is needed also in the shock tube. Nevertheless, the ideas of moving shock will reduced the work for many student of compressible flow. For example solving homework problem from other text books became either just two mouse clicks away or just looking at that the tables in this book. I also got request from a India to write the interface for Microsoft. I am sorry will not be entertaining work for non Linux/Unix systems, especially for Microsoft. If one want to use the software engine it is okay and permitted by the license of this work. The download to this mount is over 25,000.

Version 0.4.2 It was surprising to ﬁnd that over 14,000 downloaded and is encouraging to receive over 200 thank you eMail (only one from U.S.A./Arizona) and some other reactions.

xxxv xxxvi LIST OF TABLES This textbook has sections which are cutting edge research8. The additions of this version focus mainly on the oblique shock and related issues as results of questions and reactions on this topic. However, most readers reached to www.potto.org by searching for either terms “Rayleigh flow” (107) and “Fanno flow” ((93). If the total combined variation search of terms “Fanno” and “Rayleigh” (mostly through google) is accounted, it reaches to about 30% (2011). This indicates that these topics are highly is demanded and not many concerned with the shock phenomena as this author believed and expected. Thus, most additions of the next version will be concentrated on Fanno flow and Rayleigh flow. The only exception is the addition to Taylor–Maccoll flow (axisymmetricale conical flow) in Prandtl -Meyer function (currently in a note form). Furthermore, the questions that appear on the net will guide this author on what is really need to be in a compressible flow book. At this time, several questions were about compressibility factor and two phase flow in Fanno flow and other kind of flow models. The other questions that appeared related two phase and connecting several chambers to each other. Also, an individual asked whether this author intended to write about the unsteady section, and hopefully it will be near future.

Version 0.4 Since the last version (0.3) several individuals sent me remarks and suggestions. In the introductory chapter, extensive description of the compressible flow history was written. In the chapter on speed of sound, the two phase aspects were added. The isothermal nozzle was combined with the isentropic chapter. Some examples were added to the normal shock chapter. The fifth chapter deals now with normal shock in variable area ducts. The sixth chapter deals with external forces fields. The chapter about oblique shock was added and it contains the analytical solution. At this stage, the connection between Prandtl–Meyer flow and oblique is an note form. The a brief chapter on Prandtl–Meyer flow was added.

Version 0.3 In the traditional class of compressible flow it is assumed that the students will be aerospace engineers or dealing mostly with construction of airplanes and turbomachinery. This premise should not be assumed. This assumption drives students from other fields away from this knowledge. This knowledge should be spread to other fields because it needed there as well. This “rejection” is especially true when students feel that they have to go through a “shock wave” in their understanding. This book is the second book in the series of POTTO project books. POTTO project books are open content textbooks. The reason the topic of Com-

8A reader asked this author to examine a paper on Triple Shock Entropy Theorem and Its Conse- quences by Le Roy F. Henderson and Ralph Menikoff. This led to comparison between maximum to ideal gas model to more general model. VERSION 0.3 xxxvii pressible Flow was chosen, while relatively simple topics like fundamentals of strength of material were delayed, is because of the realization that manufacture engineering simply lacks fundamental knowledge in this area and thus produces faulty designs and understanding of major processes. Unfortunately, the undersigned observed that many researchers who are dealing with manufacturing processes are lack of understanding about fluid mechanics in general but particularly in relationship to compressible flow. In fact one of the reasons that many manufacturing jobs are moving to other countries is because of the lack of understanding of fluid mechanics in general and compressible in particular. For example, the lack of competitive advantage moves many of the die casting operations to off shore9. It is clear that an understanding of Compressible Flow is very important for areas that traditionally have ignored the knowledge of this topic10. As many instructors can recall from their time as undergraduates, there were classes during which most students had a period of confusion, and then later, when the dust settled, almost suddenly things became clear. This situation is typical also for Compressible Flow classes, especially for external compressible flow (e.g. flow around a wing, etc.). This book offers a more balanced emphasis which focuses more on internal compressible flow than the traditional classes. The internal flow topics seem to be common for the “traditional” students and students from other fields, e.g., manufacturing engineering. This book is written in the spirit of my adviser and mentor E.R.G. Eckert. Who, aside from his research activity, wrote the book that brought a revolution in the heat transfer field of education. Up to Eckert’s book, the study of heat transfer was without any dimensional analysis. He wrote his book because he realized that the dimensional analysis utilized by him and his adviser (for the post doc), Ernst Schmidt, and their colleagues, must be taught in engineering classes. His book met strong criticism in which some called to burn his book. Today, however, there is no known place in world that does not teach according to Eckert’s doctrine. It is assumed that the same kind of individuals who criticized Eckert’s work will criticize this work. This criticism will not change the future or the success of the ideas in this work. As a wise person says “don’t tell me that it is wrong, show me what is wrong”; this is the only reply. With all the above, it must be emphasized that this book will not revolutionize the field even though considerable new materials that have never been published are included. Instead, it will provide a new emphasis and new angle to Gas Dynamics. Compressible flow is essentially different from incompressible flow in mainly two respects: discontinuity (shock wave) and choked flow. The other issues, while important, are not that crucial to the understanding of the unique phenomena of compressible flow. These unique issues of compressible flow are to be emphasized and shown. Their applicability to real world processes is to be

9Please read the undersigned’s book “Fundamentals of Die Casting Design,” which demonstrates how ridiculous design and research can be. 10The fundamental misunderstanding of choking results in poor models (research) in the area of die casting, which in turn results in many bankrupt companies and the movement of the die casting industry to offshore. xxxviii LIST OF TABLES demonstrated11. The book is organized into several chapters which, as a traditional textbook, deals with a basic introduction of thermodynamics concepts (under construction). The second chapter deals with speed of sound. The third chapter provides the first example of choked flow (isentropic flow in a variable area). The fourth chapter deals with a simple case of discontinuity (a simple shock wave in a nozzle). The next chapter is dealing with isothermal flow with and without external forces (the moving of the choking point), again under construction. The next three chapters are dealing with three models of choked flow: Isothermal flow12, Fanno flow and Rayleigh flow. First, the Isothermal flow is introduced because of the relative ease of the analytical treatment. Isothermal flow provides useful tools for the pipe systems design. These chapters are presented almost independently. Every chapter can be “ripped” out and printed independently. The topics of filling and evacuating of gaseous chambers are presented, normally missed from traditional textbooks. There are two advanced topics which included here: oblique shock wave, and properties change effects (ideal gases and real gases) (under construction). In the oblique shock, for the first time analytical solution is presented, which is excellent tool to explain the strong, weak and unrealistic shocks. The chapter on one-dimensional unsteady state, is currently under construction. The last chapter deals with the computer program, Gas Dynamics Cal- culator (CDC-POTTO). The program design and how to use the program are described (briefly). Discussions on the flow around bodies (wing, etc), and Prandtl–Meyer expansion will be included only after the gamma version unless someone will provide discussion(s) (a skeleton) on these topics. It is hoped that this book will serve the purposes that was envisioned for the book. It is further hoped that others will contribute to this book and find additional use for this book and enclosed software.

11If you have better and different examples or presentations you are welcome to submit them. 12It is suggested to referred to this model as Shapiro ﬂow How This Book Was Written

This book started because I needed an explanation for manufacturing engineers. Apparently many manufacturing engineers and even some researchers in manufacturing engineering were lack of understanding about fluid mechanics in particularly about compressible flow. Therefore, I wrote to myself some notes and I converted one of the note to a chapter in my first book, “Fundamentals Of Die Casting Design.” Later, I realized that people need down to earth book about compressible flow and this book was born. The free/open content of the book was created because the realization that open content accelerated the creation of books and reaction to the corruption of the court implementing the copyright law by manufacturing facts and laws. It was farther extended by the allegation of free market and yet the academic education cost is sky rocketing without a real reason and real competition. There is no reason why a text book which cost leas than 10$ to publish/produce will cost about 150 dollars. If a community will pull together, the best books can be created. Anyone can be part of it. For example, even my 10 years old son, Eliezer made me change the chapter on isothermal flow. He made me realized that the common approach to supersonic branch of isothermal as non–existent is the wrong approach. It should be included because this section provides the explanation and direction on what Fanno flow model will approach if heat transfer is taken into account13. I realized that books in compressible flow are written in a form that is hard for non fluid mechanic engineer to understand. Therefore, this book is designed to be in such form that is easy to understand. I wrote notes and asked myself what materials should be included in such a book so when I provide consultation to a company, I do not need to explain the fundamentals. Therefore, there are some chapters in this book which are original materials never published before. The presentation of some of the chapters is different from other books. The book

13Still in untyped note form.

xxxix xl LIST OF TABLES does not provide the old style graphical solution methods yet provide the graphical explanation of things. Of course, this book was written on Linux (MicrosoftLess book). This book was written using the vim editor for editing (sorry never was able to be com- fortable with emacs). The graphics were done by TGIF, the best graphic program that this author experienced so far. The old ﬁgures where done by grap (part the old Troff). Unfortunately, I did not have any access to grap and switched to Grace. Grace is a problematic program but is the best I have found. The spell checking was done by gaspell, a program that cannot be used on new system and I had to keep my old Linux to make it work14. I hope someone will write a new spell check so I can switch to a new system. The ﬁgure in cover page was created by Michael Petschauer, graphic designer, and is open/free content copyright by him ( happy [email protected]).

14If you would like to to help me to write a new spell check user interface, please contact me. About Gas Dynamics Calculator

Gas Dynamic Calculator, (Potto–GDC) was created to generate various tables for the book either at end the chapters or for the exercises. This calculator was given to several individuals and they found Potto–GDC to be very useful. So, I decided to include Potto–GDC to the book. Initially, the Potto-GDC was many small programs for specific tasks. For example, the stagnation table was one such program. Later, the code became a new program to find the root of something between the values of the tables e.g. 4fL finding parameters for a given D . At that stage, the program changed to contain a primitive interface to provide parameters to carry out the proper calculations. Yet, then, every flow model was a different program. When it become cumbersome to handle several programs, the author utilized the object oriented feature of C++ and assigned functions to the common tasks to a base class and the specific applications to the derived classes. Later, a need to intermediate stage of tube flow model (the PipeFlow class) was created and new classes were created. The graphical interface was created only after the engine was written. The graphical interface was written to provide a filter for the unfamiliar user. It also remove the need to recompile the code everytime.

Version 4.3

This version add several feature among them is the shock dynamics calculation with the iteration. The last freature is good for homework either for the students or the instroctors.

xli xlii LIST OF TABLES Version 4.1.7 Version 4.1.7 had several bug ﬁxes and add two angle calculations to the oblique shock. Change the logtable to tabular environment for short tables. Preface

"In the beginning, the POTTO project was without form, and void; and emptiness was upon the face of the bits and files. And the Fingers of the Author moved upon the face of the keyboard. And the Author said, Let there be words, and there were words." 15.

This book, Fundamentals of Compressible Flow, describes the fundamentals of compressible flow phenomena for engineers and others. This book is designed to replace the book(s) or instructor’s notes for the compressible flow in (mostly) undergraduate classes for engineering/science students. It is hoped that the book could be used as a reference book for people who have at least some knowledge of the basics of fundamental fluid mechanics, and basic science such as calculus, physics, etc. It is hoped that the computer program enclosed in the book will take on a life of its own and develop into an open content or source project. The structure of this book is such that many of the chapters could be usable independently. For example, if you need information about, say, Fanno flow, you can read just chapter 9. I hope this makes the book easier to use as a reference manual. However, this manuscript is first and foremost a textbook, and secondly a reference manual only as a lucky coincidence. I have tried to describe why the theories are the way they are, rather than just listing “seven easy steps” for each task. This means that a lot of information is presented which is not necessary for everyone. These explanations have been marked as such and can be skipped.16 Reading everything will, naturally, increase your understanding of the fundamentals of compressible fluid flow. This book is written and maintained on a volunteer basis. Like all volunteer work, there is a limit on how much effort I was able to put into the book and its organization. Moreover, due to the fact that English is my third language and time limitations, the explanations are not as good as if I had a few years to perfect them. Nevertheless, I believe professionals working in many engineering

15To the power and glory of the mighty God. This book is only to explain his power. 16At the present, the book is not well organized. You have to remember that this book is a work in progress.

xliii xliv LIST OF TABLES ﬁelds will beneﬁt from this information. This book contains many original models, and explanations never published before. I have left some issues which have unsatisfactory explanations in the book, marked with a Mata mark. I hope to improve or to add to these areas in the near future. Furthermore, I hope that many others will participate of this project and will contribute to this book (even small contributions such as providing examples or editing mistakes are needed). I have tried to make this text of the highest quality possible and am interested in your comments and ideas on how to make it better. Incorrect language, errors, ideas for new areas to cover, rewritten sections, more fundamental material, more mathematics (or less mathematics); I am interested in it all. If you want to be involved in the editing, graphic design, or proofreading, please drop me a line. You may contact me via Email at “[email protected]”. Naturally, this book contains material that never was published before. This material never went through a peer review. While peer review and publication in a professional publication is excellent idea in theory. In practice, this process leaves a large room to blockage of novel ideas and plagiarism. If you would like be “peer reviews” or critic to my new ideas please send me your idea(s). Even reaction/comments from individuals like David Marshall17 Several people have helped me with this book, directly or indirectly. I would like to especially thank to my adviser, Dr. E. R. G. Eckert, whose work was the inspiration for this book. I also would like to thank Amy Ross for her advice ideas, and assistance. The symbol META was added to provide typographical conventions to blurb as needed. This is mostly for the author’s purposes and also for your amuse- ment. There are also notes in the margin, but those are solely for the author’s purposes, ignore them please. They will be removed gradually as the version number advances. I encourage anyone with a penchant for writing, editing, graphic ability, LATEX knowledge, and material knowledge and a desire to provide open content textbooks and to improve them to join me in this project. If you have Internet e-mail access, you can contact me at “[email protected]”.

17Dr. Marshall wrote to this author that the author should review other people work before he write any thing new (well, literature review is always good?). Over ten individuals wrote me about this letter. I am asking from everyone to assume that his reaction was innocent one. While his comment looks like unpleasant reaction, it brought or cause the expansion the oblique shock chapter. However, other email that imply that someone will take care of this author aren’t appreciated. To Do List and Road Map

This book is not complete and probably never will be completed. There will always new problems to add or to polish the explanations or include more new materials. Also issues that associated with the book like the software has to be improved. It is hoped the changes in TEX and LATEX related to this book in future will be minimal and minor. It is hoped that the style file will be converged to the final form rapidly. Nevertheless, there are specific issues which are on the “table” and they are described herein. At this stage, several chapters are missing. The effects of the deviations from the ideal gas model on the properties should be included. Further topics related to non-ideal gas such as steam and various freons are in the process of being added to this book especially in relationship to Fanno flow. One of the virtue of this book lay in the fact that it contains a software that is extensible. For example, the Fanno module can be extended to include effects of real gases. This part will be incorporated in the future hopefully with the help of others. Specific missing parts from every chapters are discussed below. These omissions, mistakes, approach problems are sometime appears in the book under the Meta simple like this Meta sample this part. Meta End Questions/problems appear as a marginal note. On occasions a footnote was used to point out for a need of improvement. You are always welcome to add a new material: problem, question, illustration or photo of experiment. Material can

xlv xlvi LIST OF TABLES be further illuminate. Additional material can be provided to give a different angle on the issue at hand.

Speed of Sound Discussion about the movement in medium with variation in speed of sound. This concept in relation of the wind tunnel and atmosphere with varied density and temperature. Mixed gases and liquids. More problems in relationship to two phase. Speed of sound in wet steam.

Stagnation effects Extend the applicability with examples Cp as a function of temperature (deviation of ideal gas model) “real gas”’ like water vapor History – on the teaching (for example when the concept of stagnation was ﬁrst taught.

Nozzle The effect of external forces (add problems). Real gases effects (only temperature effects) Flow with “tabulated gases” calculations Phase change and two phase flow (multi choking points) effects (after 1.0 version). The dimensional analysis of the flow when the flow can be considered as isothermal. The combined effects of isentropic nozzle with heat transfer (especially with relationship to the program.).

Normal Shock Extend the partialy (open/close) moving shock theory. Provide more examples on the preveious topic Shock in real gases like water vapor Shock in (partially) two phase gases like air with dust particals

Isothermal Flow Classiﬁcation of Problems Comparison of results with Fanno ﬂow Pipes Network calculations. VERSION 4.1.7 xlvii Fanno Flow More examples: various categories Some improvement on the software (clean up) Real gas effects (compressible factor) Tablated gas

Rayleigh Flow To mature the chapter: discussion on the “dark” corners of this model. Provide discussion on variations of the effecting parameters. Examples: provide categorization

Evacuation and filling semi rigid Chambers To construct the Rayleigh flow in the tube (thermal chocking) Energy equation (non isentropic process) Examples classifications Software (converting the FORTRAN program to c++)

Evacuating and ﬁlling chambers under external forces Comparison with chemical reaction case Energy equation (non isentropic process) Examples Software transformation from FORTRAN to c++. The FORTRAN version will not be included.

Oblique Shock Add application to design problems Real Gas effects

Prandtl–Meyer The limitations (Prandtl-Meyer). Application Marcell–Taylor (from the notes) Examples

Transient problem xlviii LIST OF TABLES CHAPTER 1

Introduction

1.1 What is Compressible Flow ? This book deals with an introduction1 to the flow of compressible substances (gases). The main difference between compressible flow and almost incompressible flow is not the fact that compressibility has to be considered. Rather, the difference is in two phenomena that do not exist in incompressible flow2. The first phenomenon is the very sharp discontinuity (jump) in the flow in properties. The second phenomenon is the choking of the flow. Choking is when downstream variations don’t effect the flow3. Though choking occurs in certain pipe flows in astronomy, there also are situations of choking in general (external) flow4. Choking is referred to as the situation where downstream conditions, which are beyond a critical value(s), doesn’t affect the flow. The shock wave and choking are not intuitive for most people. However, one has to realize that intuition is really a condition where one uses his past expe- riences to predict other situations. Here one has to learn to use his intuition as a tool for future use. Thus, not only aeronautic engineers, but other engineers, and even manufacturing engineers will be able use this “intuition” in design and even research.

1This book gradually sliding to include more material that isn’t so introductory. But attempt is made to present the material in introductory level. 2It can be argued that in open channel ﬂow there is a hydraulic jump (discontinuity) and in some ranges no effect of downstream conditions on the ﬂow. However, the uniqueness of the phenomena in the gas dynamics provides spectacular situations of a limited length (see Fanno model) and thermal choking, etc. Further, there is no equivalent to oblique shock wave. Thus, this richness is unique to gas dynamics. 3The thermal choking is somewhat different but similarity exists. 4This book is intended for engineers and therefore a discussion about astronomical conditions isn’t presented.

1 2 CHAPTER 1. INTRODUCTION 1.2 Why Compressible Flow is Important? Compressible flow appears in many natural and many technological processes. Compressible flow deals with more than air, including steam, natural gas, nitrogen and helium, etc. For instance, the flow of natural gas in a pipe system, a common method of heating in the u.s., should be considered a compressible flow. These processes include the flow of gas in the exhaust system of an internal combustion engine, and also gas turbine, a problem that led to the Fanno flow model. The above flows that were mentioned are called internal flows. Compressible flow also includes flow around bodies such as the wings of an airplane, and is considered an external flow. These processes include situations not expected to have a compressible flow, such as manufacturing process such as the die casting, injection molding. The die casting process is a process in which liquid metal, mostly aluminum, is injected into a mold to obtain a near final shape. The air is displaced by the liquid metal in a very rapid manner, in a matter of milliseconds, therefore the compressibility has to be taken into account. Clearly, Aero Engineers are not the only ones who have to deal with some aspect of compressible flow. For manufacturing engineers there are many situations where the compressibility or compressible flow understating is essential for adequate design. For instance, the control engineers who are using pneumatic systems use compressed substances. The cooling of some manufacturing systems and design of refrigeration systems also utilizes compressed air flow knowledge. Some aspects of these systems require consideration of the unique phenomena of compressible flow. Traditionally, most gas dynamics (compressible flow) classes deal mostly with shock waves and external flow and briefly teach Fanno flows and Rayleigh flows (two kind of choking flows). There are very few courses that deal with isothermal flow. In fact, many books on compressible flow ignore the isothermal flow5. In this book, a greater emphasis is on the internal flow. This doesn’t in any way meant that the important topics such as shock wave and oblique shock wave should be neglected. This book contains several chapters which deal with external flow as well.

1.3 Historical Background In writing this book it became clear that there is more unknown and unwritten about the history of compressible fluid than known. While there are excellent books about the history of fluid mechanics (hydraulic) see for example book by Rouse6. There are numerous sources dealing with the history of flight and airplanes (aeronau-

5Any search on the web on classes of compressible flow will show this fact and the undersigned can testify that this was true in his first class as a student of compressible flow. 6Hunter Rouse and Simon Inc, History of Hydraulics (Iowa City: Institute of Hydraulic Research, 1957) 1.3. HISTORICAL BACKGROUND 3 tic)7. Aeronautics is an overlapping part of compressible flow, however these two fields are different. For example, the Fanno flow and isothermal flow, which are the core of gas dynamics, are not part of aerodynamics. Possible reasons for the lack of written documentation are one, a large part of this knowledge is relatively new, and two, for many early contributors this topic was a side issue. In fact, only one contributor of the three main models of internal compressible flow (Isothermal, Fanno, Rayleigh) was described by any text book. This was Lord Rayleigh, for whom the Rayleigh flow was named. The other two models were, to the undersigned, unknown. Furthermore, this author did not find any reference to isothermal flow model earlier to Shapiro’s book. There is no book8 that describes the history of these models. For instance, the question, who was Fanno, and when did he live, could not be answered by any of the undersigned’s colleagues in University of Minnesota or elsewhere. At this stage there are more questions about the history of compressible flow needing to be answered. Sometimes, these questions will appear in a section with a title but without text or with only a little text. Sometimes, they will appear in a footnote like this9. For example, it is obvious that Shapiro published the erroneous conclusion that all the chocking occurred at M = 1 in his article which contradicts his isothermal model. Additional example, who was the first to “conclude” the “all” the chocking occurs at M = 1? Is it Shapiro? Originally, there was no idea that there are special effects and phenomena of compressible flow. Some researchers even have suggested that compressibility can be “swallowed” into the ideal flow (Euler’s equation’s flow is sometimes referred to as ideal flow). Even before Prandtl’s idea of boundary layer appeared, the significant and importance of compressibility emerged. In the first half of nineteen century there was little realization that the compressibility is important because there were very little applications (if any) that required the understanding of this phenomenon. As there were no motivations to investigate the shock wave or choked flow both were treated as the same, taking compressible flow as if it were incompressible flow. It must be noted that researchers were interested in the speed of sound even long before applications and knowledge could demand any utilization. The research and interest in the speed of sound was a purely academic interest. The early application in which compressibility has a major effect was with fire arms. The technological improvements in fire arms led to a gun capable of shooting bullets at speeds approaching to the speed of sound. Thus, researchers were aware that the speed of sound is some kind of limit. In the second half of the nineteen century, Mach and Fliegner “stumbled” over the shock wave and choking, respectively. Mach observed shock and Fliegner

7Anderson, J. D., Jr. 1997. A History of Aerodynamics: And Its Impact on Flying Machines, Cam- bridge University Press, Cambridge, England. 8The only remark found about Fanno flow that it was taken from the Fanno Master thesis by his adviser. Here is a challenge: find any book describing the history of the Fanno model. 9Who developed the isothermal model? The research so far leads to Shapiro. Perhaps this flow should be named after the Shapiro. Is there any earlier reference to this model? 4 CHAPTER 1. INTRODUCTION measured the choking but theoretical science did not provide explanation for it (or was award that there is an explanation for it.). In the twentieth century the flight industry became the pushing force. Un- derstandably, aerospace engineering played a significant role in the development of this knowledge. Giants like Prandtl and his students like Van Karman, as well as others like Shapiro, dominated the field. During that time, the modern basic classes became “solidified.” Contributions by researchers and educators from other fields were not as dominant and significant, so almost all text books in this field are writ- To add history from the work. ten from an aerodynamic prospective. Topics that should be included in this history review but that are not yet added to this section are as follows: Multi Phase flow, cap- illary flow and phase change. 1.3.1 Early Developments

The compressible flow is a subset of fluid mechanics/hydraulics and therefore the knowledge development followed the understanding of incompressible flow. Early contributors were motivated from a purely intellectual curiosity, while most later contributions were driven by necessity. As a result, for a long time the question of the speed of sound was bounced around.

Speed of Sound

The idea that there is a speed of sound and that it can be measured is a major achievement. A possible explanation to this discovery lies in the fact that mother nature exhibits in every thunder storm the difference between the speed of light and the speed of sound. There is no clear evidence as to who came up with this concept, but some attribute it to Galileo Galilei: 166x. Galileo, an Italian scientist, was one of the earliest contributors to our understanding of sound. Dealing with the difference between the two speeds (light, sound) was a major part of Galileo’s work. However, once there was a realization that sound can be measured, people found that sound travels in different speeds through different mediums. The early approach to the speed of sound was by the measuring of the speed of sound. Other milestones in the speed of sound understanding development were by Leonardo Da Vinci, who discovered that sound travels in waves (1500). Marin Mersenne was the first to measure the speed of sound in air (1640). Robert Boyle discovered that sound waves must travel in a medium (1660) and this lead to the concept that sound is a pressure change. Newton was the first to formulate a relationship between the speed of sound in gases by relating the density and compressibility in a medium (by assuming isothermal process). Newton’s equation is missing the heat ratio, k (late 1660’s). Maxwell was the first to derive the speed of sound for gas as c = √kRT from particles (statistical) mechanics. Therefore some referred to coefficient √k as Maxwell’s coefficient. 1.3. HISTORICAL BACKGROUND 5 1.3.2 The shock wave puzzle Here is where the politics of science was a major obstacle to achieving an ad- vancement10. In the early 18xx, conservation of energy was a concept that was applied only to mechanical energy. On the other side, a different group of scientists dealt with calorimetry (internal energy). It was easier to publish articles about the second law of thermodynamics than to convince anyone of the first law of thermodynamics. Neither of these groups would agree to “merge” or “relinquish” control of their “territory” to the other. It took about a century to establish the first law11. At first, Poisson found a “solution” to the Euler’s equations with certain boundary conditions which required discontinuity12 which had obtained an implicit form in 1808. Poisson showed that solutions could approach a discontinuity by using conservation of mass and momentum. He had then correctly derived the jump conditions that discontinuous solutions must satisfy. Later, Challis had noticed contradictions concerning some solutions of the equations of compressible gas dynamics13. Again the “jumping” conditions were redeveloped by two different researchers independently: Stokes and Riemann. Riemann, in his 1860 thesis, was not sure whether or not discontinuity is only a mathematical creature or a real creature. Stokes in 1848 retreated from his work and wrote an apology on his “mistake.”14 Stokes was convinced by Lord Rayleigh and Lord Kelvin that he was mistaken on the grounds that energy is conserved (not realizing the concept of internal energy). At this stage some experimental evidence was needed. Ernst Mach studied several fields in physics and also studied philosophy. He was mostly interested in experimental physics. The major breakthrough in the understanding of compressible flow came when Ernest Mach “stumbled” over the discontinuity. It is widely believed that Mach had done his research as purely intellectual research. His research centered on optic aspects which lead him to study acoustic and therefore supersonic flow (high speed, since no Mach number was known at that time). However, it is logical to believe that his interest had risen due to the need to achieve

10Amazingly, science is full of many stories of conflicts and disputes. Aside from the conflicts of scientists with the Catholic Church and Muslim religion, perhaps the most famous is that of Newton’s netscaping (stealing and embracing) Leibniz[’s] invention of calculus. There are even conflicts from not giving enough credit, like Moody not giving the due credit to Rouse. Even the undersigned encountered individuals who have tried to ride on his work. The other kind of problem is “hijacking” by a sector. Even on this subject, the Aeronautic sector “took over” gas dynamics as did the emphasis on mathematics like perturbations methods or asymptotic expansions instead on the physical phenomena. Major material like Fanno flow isn’t taught in many classes, while many of the mathematical techniques are currently practiced. So, these problems are more common than one might be expected. 11This recognition of the first law is today the most “obvious” for engineering students. Yet for many it was still debatable up to the middle of the nineteen century. 12Simeon´ Denis Poisson, French mathematician, 1781-1840 worked in Paris, France. ”M’emoire sur la th’eorie du son,” J. Ec. Polytech. 14 (1808), 319-392. From Classic Papers in Shock Compression Science, 3-65, High-press. Shock Compression Condens. Matter, Springer, New York, 1998. 13James Challis, English Astronomer, 1803-1882. worked at Cambridge, England UK. ”On the velocity of sound,” Philos. Mag. XXXII (1848), 494-499 14Stokes George Gabriel Sir, Mathematical and Physical Papers, Reprinted from the original journals and transactions, with additional notes by the author. Cambridge, University Press, 1880-1905. 6 CHAPTER 1. INTRODUCTION powerful/long–distance shooting rifles/guns. At that time many inventions dealt with machine guns which were able to shoot more bullets per minute. At the time, one anecdotal story suggests a way to make money by inventing a better killing machine for the Europeans. While the machine gun turned out to be a good killing machine, defense techniques started to appear such as sand bags. A need for bullets that could travel faster to overcome these obstacles was created. There- fore, Mach’s paper from 1876 deals with the flow around bullets. Nevertheless, no known15 equations or explanations resulted from these experiments. Mach used his knowledge in Optics to study the flow around bullets. What makes Mach’s achievement all the more remarkable was the technique he used to take the historic photograph: He employed an innovative approach called the shadowgraph. He was the first to photograph the shock wave. In his paper discussing ”Photographische Fixierung der durch Projektile in der Luft eingeleiten Vorgange” he showed a picture of a shock wave (see Figure 1.7). He utilized the variations of the air density to clearly show shock line at the front of the bullet. Mach had good understanding of the fundamentals of supersonic flow and the effects on bullet movement (supersonic flow). Mach’s paper from 1876 demonstrated shock wave (discontinuity) and suggested the importance of the ratio of the velocity to the speed of sound. He also observed the existence of a conical shock wave (oblique shock wave). Mach’s contributions can be summarized as providing an experimental proof to discontinuity. He further showed that the discontinuity occurs at M = 1 and realized that the velocity ratio (Mach number), and not the velocity, is the important parameter in the study of the compressible flow. Thus, he brought confidence to the theoreticians to publish their studies. While Mach proved shock wave and oblique shock wave existence, he was not able to analyze it (neither was he aware of Poisson’s work or the works of others.). Back to the pencil and paper, the jump conditions were redeveloped and now named after Rankine16 and Hugoniot17. Rankine and Hugoniot, redeveloped independently the equation that governs the relationship of the shock wave. Shock was assumed to be one dimensional and mass, momentum, and energy equa- tions18 lead to a solution which ties the upstream and downstream properties. What they could not prove or find was that shock occurs only when upstream is supersonic, i.e., direction of the flow. Later, others expanded Rankine-Hugoniot’s

15The words “no known” refer to the undersigned. It is possible that some insight was developed but none of the documents that were reviewed revealed it to the undersigned. 16William John Macquorn Rankine, Scottish engineer, 1820-1872. He worked in Glasgow, Scotland UK. ”On the thermodynamic theory of waves of finite longitudinal disturbance,” Philos. Trans. 160 (1870), part II, 277-288. Classic papers in shock compression science, 133-147, High-press. Shock Compression Condens. Matter, Springer, New York, 1998 17Pierre Henri Hugoniot, French engineer, 1851-1887. ”Sur la propagation du mouvement dans les corps et sp’ecialement dans les gaz parfaits, I, II” J. Ec. Polytech. 57 (1887), 3-97, 58 (1889), 1-125. Classic papers in shock compression science, 161-243, 245-358, High-press. Shock Compression Condens. Matter, Springer, New York, 1998 18Today it is well established that shock has three dimensions but small sections can be treated as one dimensional. 1.3. HISTORICAL BACKGROUND 7 conditions to a more general form19. Here, the second law has been around for over 40 years and yet the significance of it was not was well established. Thus, it took over 50 years for Prandtl to arrive at and to demonstrate that the shock has only one direction20. Today this equation/condition is known as Prandtl’s equation or condition (1908). In fact Prandtl is the one who introduced the name of Rankine-Hugoniot’s conditions not aware of the earlier developments of this condition. Theodor Meyer (Prandtl’s student) derived the conditions for oblique shock in 190821 as a byproduct of the expansion work. It was probably later that Stodola (Fanno’s adviser) realized that the shock is the intersection of the Fanno line with the Rayleigh line. Yet, the supersonic branch is missing from his understanding (see Figure (1.1)). In fact, Stodola suggested the graphical solution utilizing the Fanno line. The fact that the conditions and direction were known did not bring the solution to the equations. The “last nail” of understanding was put by Landau, a Jewish scientist who worked in Moscow University in the 1960’s during the Com- Fig. 1.1: The shock as connection of Fanno and munist regimes. A solution Rayleigh lines after Stodola, Steam and Gas was found by Landau & Lifshitz Turbine and expanded by Kolosnitsyn & Stanyukovich (1984). to be add to oblique shock chapter. Since early in the 1950s the analytical relationships between the oblique shock, deflection angle, shock angle, and Mach number was described as impossible to obtain. There were until recently (version 0.3 of this book) several equations that tied various properties/quantities for example, the relationship between upstream Mach number and the angles. The first full analytical solution connecting the angles with upstream Mach number was published in this book version 0.3. The probable reason that analytical solution was not published because the claim

19To add discussion about the general relationships. 20Some view the work of G. I. Taylor from England as the proof (of course utilizing the second law) 21Theodor Meyer in Mitteil. ub¨ . Forsch-Arb. Berlin, 1908, No. 62, page 62. 8 CHAPTER 1. INTRODUCTION in the famous report of NACA 1135 that explicit analytical solution isn’t possible22. The question whether the oblique shock is stable or which root is stable was daunting since the early discovery that there are more than one possible solution. It is amazing that early research concluded that only the weak solution is possible or stable as opposed to the reality. The first that attempt this question where in 1931 by Epstein23. His analysis was based on Hamilton’s principle when he ignore the boundary condition. The results of that analysis was that strong shock is unstable. The researchers understood that flow after a strong shock was governed by elliptic equation while the flow after a weak shock was governed by hyperbolic equations. This difference probably results in not recognizing that The boundary conditions play an important role in the stability of the shock24. In fact analysis based on Hamilton’s principle isn’t suitable for stability because entropy creation was recognized 1955 by Herivel25. Carrier26 was first to recognize that strong and weak shocks stable. If fact the confusion on this issue was persistent until now. Even all books that were published recently claimed that no strong shock was ever observed in flow around cone (Taylor–Maccoll flow). In fact, even this author sinned in this erroneous conclusion. The real question isn’t if they exist rather under what conditions these shocks exist which was suggested by Courant and Friedrichs in their book “Supersonic Flow and Shock Waves,” published by Interscience Publishers, Inc. New York, 1948, p. 317. The effect of real gases was investigated very early since steam was used move turbines. In general the mathematical treatment was left to numerical investigation and there is relatively very little known on the difference between ideal gas model and real gas. For example, recently, Henderson and Menikoff27 dealt with only the procedure to find the maximum of oblique shock, but no comparison between real gases and ideal gas is offered there.

22Since writing this book, several individuals point out that a solution was found in book “Analytical Fluid Dynamics” by Emanuel, George, second edition, December 2000 (US$ 124.90). That solution is based on a transformation of sin θ to tan β. It is interesting that transformation result in one of root being negative. While the actual solution all the roots are real and positive for the attached shock. The presentation was missing the condition for the detachment or point where the model collapse. But more surprisingly, similar analysis was published by Briggs, J. “Comment on Calculation of Oblique shock waves,” AIAA Journal Vol 2, No 5 p. 974, 1963. Hence, Emanuel’s partial solution just redone 36 years work (how many times works have to be redone in this field). In a way, part of analysis of this book is also redoing old work. Yet, what is new in this work is completeness of all the three roots and the analytical condition for detached shock and breaking of the model. 23Epstein, P. S., “On the air resistance of Projectiles,” Proceedings of the National Academy of Sci- ence, Vol. 17, 1931, pp. 532-547. 24In study this issue this author realized only after examining a colleague experimental Picture 14.4 that it was clear that the Normal shock along with strong shock and weak shock “live” together peacefully and in stable conditions. 25Herivel, J. F., “The Derivation of The Equations of Motion On an Ideal Fluid by Hamilton’s Principle,,” Proceedings of the Cambridge philosophical society, Vol. 51, Pt. 2, 1955, pp. 344-349. 26Carrier, G.F., “On the Stability of the supersonic Flows Past as a Wedge,” Quarterly of Applied Mathematics, Vol. 6, 1949, pp. 367–378. 27Henderson and Menikoff, ”Triple Shock Entropy Theorem,” Journal of Fluid Mechanics 366 (1998) pp. 179–210. 1.3. HISTORICAL BACKGROUND 9 The moving shock and shock tube were study even before World War Two. The realization that in most cases the moving shock can be analyzed as steady state since it approaches semi steady state can be traced early of 1940’s. Up to this version 0.4.3 of this book (as far it is known, this book is first to publish this tables), trial and error method was the only method to solve this problem. Only after the dimensionless presentation of the problem and the construction of the moving shock table the problem became trivial. Later, an explicit analytical solution for shock a head of piston movement (special case of open valve) was originally published in this book for the first time.

1.3.3 Choking Flow

The choking problem is almost unique to gas dynamics and has many different forms. Choking wasn’t clearly to be observed, even when researcher stumbled over it. No one was looking for or expecting the choking to occur, and when it was found the significance of the choking phenomenon was not clear. The first experimental choking phenomenon was discovered by Fliegner’s experiments which were conducted some time in the middle of 186x28 on air flow through a converging nozzle. As a result deLavel’s nozzle was invented by Carl Gustaf Patrik Fig. 1.2: The schematic of deLavel’s turbine af- de Laval in 1882 and first successful ter Stodola, Steam and Gas Turbine operation by another inventor (Curtis) 1896 used in steam turbine. Yet, there was no realization that the flow is choked just that the flow moves faster than speed of sound. The introduction of the steam engine and other thermodynamics cycles led to the choking problem. The problem was introduced because people wanted to increase the output of the Engine by increasing the flames (larger heat transfer or larger energy) which failed, leading to the study and development of Rayleigh flow. According the thermodynamics theory (various cycles) the larger heat supply for a given temperature difference (larger higher temperature) the larger the output, but after a certain point it did matter (because the steam was choked). The first to discover (try to explain) the choking phenomenon was Rayleigh29.

28Fliegner Schweizer Bauztg., Vol 31 1898, p. 68–72. The theoretical first work on this issue was done by Zeuner, “Theorie die Turbinen,” Leipzig 1899, page 268 f. 29Rayleigh was the first to develop the model that bears his name. It is likely that others had noticed that flow is choked, but did not produce any model or conduct successful experimental work. 10 CHAPTER 1. INTRODUCTION After the introduction of the deLavel’s converging–diverging nozzle theoretical work was started by Zeuner30. Later continue by Prandtl’s group31 starting 1904. In 1908 Meyer has extend this work to make two dimensional calcula- tions32. Experimental work by Parenty33 and others measured the pressure along the converging-diverging nozzle.

It was commonly believed34 that the choking occurs only at M = 1. The first one to analyzed that choking occurs at 1/√k for isothermal flow was Shapiro (195x). It is so strange that a giant like Shapiro did not realize his model on isothermal contradict his conclusion from his own famous paper. Later Romer at el extended it to isothermal variable area flow (1955). In this book, this author adapts E.R.G. Ecert’s idea of dimensionless parameters control which determines where the reality lay between the two extremes. Recently this concept was proposed (not explicitly) by Dutton and Converdill (1997)35. Namely, in many cases the reality is somewhere between the adiabatic and the isothermal flow. The actual results will be determined by the modified Eckert number to which model they are closer.

30Zeuner, “Theorie der Turbinen, Leipzig 1899 page 268 f. 31Some of the publications were not named after Prandtl but rather by his students like Meyer, Theodor. In the literature appeared reference to article by Lorenz in the Physik Zeitshr., as if in 1904. Perhaps, there are also other works that this author did not come crossed. 32Meyer, Th., Uber¨ zweidimensionals Bewegungsvordange eines Gases, Dissertation 1907, er- schienen in den Mitteilungen uber¨ Forsch.-Arb. Ing.-Wes. heft 62, Berlin 1908. 33Parenty, Comptes R. Paris, Vol. 113, 116, 119; Ann. Chim. Phys. Vol. 8. 8 1896, Vol 12, 1897. 34The personal experience of this undersigned shows that even instructors of Gas Dynamics are not aware that the chocking occurs at different Mach number and depends on the model. 35These researchers demonstrate results between two extremes and actually proposed this idea. However, that the presentation here suggests that topic should be presented case between two extremes. 1.3. HISTORICAL BACKGROUND 11

Nozzle flow The first “wind tunnel” was not a tunnel but a rotating arm attached at the center. At the end of the arm was the object that was under observation and study. The arm’s circular motion could reach a velocity above the speed of sound at its end. Yet, in 1904 the Wright brothers demonstrated that results from the wind tunnel and spinning arm are different, due to the circular motion. As a result, the spinning arm was no longer used in testing. Be- tween the turn of the century Fig. 1.3: The measured pressure in a nozzle taken and 1947-48, when the first supersonic windfromtunnelStodolawas1927built,SteamsevanderalGasmodelsTurbinesthat explained choking at the throat have been built.

A different reason to study the converging-diverging nozzle was the Venturi meter which was used in measuring the flow rate of gases. Bendemann 36 carried experiments to study the accuracy of these flow meters and he measured and refound that the flow reaches a critical value (pressure ratio of 0.545) that creates the maximum flow rate.

There are two main models or extremes that describe the ﬂow in the nozzle: isothermal and adiabatic.

36Bendemann Mitteil uber¨ Forschungsarbeiten, Berlin, 1907, No. 37. 12 CHAPTER 1. INTRODUCTION

Nozzle flow Romer et al37analyzed the isothermal flow in a nozzle. It is remarkable that choking was found as 1/√k as opposed to one (1). In general when the model is assumed to be isothermal the choking occurs at 1/√k. The concept that the choking point can move from the throat introduced by38a researcher unknown to this author. It is very interesting that the isothermal nozzle was proposed by Romer at el 1955 (who was behind the adviser or the student?). These researchers Fig. 1.4: Flow rate as a function of the back pressure taken were the first ones to real- from Stodola 1927 Steam and Gas Turbines ized that choking can occurs at different Mach number (1/√k other then the to insert the isothermal noz- isothermal pipe. zle with external forces like gravity and to show that choking location can move depending on the direction of the force. Rayleigh Flow Rayleigh was probably39, the first to suggest a model for frictionless flow with a constant heat transfer. Rayleigh’s work was during the time when it was debatable as to whether there are two forms of energies (mechanical, thermal), even though Watt and others found and proved that they are the same. Therefore, Rayleigh looked at flow without mechanical energy transfer (friction) but only thermal energy To find where Rayleigh did transfer. In Rayleigh flow, the material reaches choking point due to heat transfer, understand that his model leads to 1/√k point flow hence term “thermally choked” is used; no additional flow can occur. and graphical representation of the flow. The 1/√k question. to insert information about the Fanno Flow detonation wave and relationship to Rayleigh line. The most important model in compressible flow was suggested by Gino Fanno in his Master’s thesis (1904). The model bears his name. Yet, according to Dr. Rudolf

38Romer, I Carl Jr., and Ali Bulent Cambel, “Analysis of Isothermal Variable Area Flow,” Aircraft Eng. vol. 27 no 322, p. 398 December 1955. 38This undersign didn’t find the actual trace to the source of proposing this effect. However, some astronomy books showing this effect in a dimensional form without mentioning the original researcher. In dimensionless form, this phenomenon produces a dimensionless number similar to Ozer number and therefor the name Ozer number adapted in this book. 39As most of the history research has shown, there is also a possibility that someone found it earlier. For example, Piosson was the first one to realize the shock wave possibility. 1.3. HISTORICAL BACKGROUND 13 Mumenthaler from UTH University, no copy of the thesis can be found in the original University and perhaps only in the personal custody of the Fanno family40. Fanno attributes the main pressure reduction to friction. Thus, flow that is domi- nantly adiabatic could be simplified and analyzed. The friction factor is the main component in the analysis as Darcy f 41 had already proposed in 1845. The arrival of the Moody diagram, which built on Hunter Rouse’s (194x) work made Darcy– Weisbach’s equation universally useful. Without the existence of the friction factor data, the Fanno model wasn’t able to produce a prediction useful for the industry. Additionally an understating of the supersonic branch of the flow was unknown (The idea of shock in tube was not raised at that time.). Shapiro organized all the material in a coherent way and made this model useful. Meta Did Fanno realize that the flow is choked? It appears at least in Stodola’s book that choking was understood in 1927 and even earlier. The choking was assumed only to be in the subsonic flow. But because the actual Fanno’s thesis is not available, the question cannot be answered yet. When was Gas Dynamics (compressible flow) as a separate class started? Did the explanation for the combination of diverging-converging nuzzle with tube for Fanno flow first appeared in Shapiro’s book? Meta End expanding model by others

Isothermal Flow

The earliest reference to isothermal flow was found in Shapiro’s Book. The model If it turned out that no one had done it before Shapiro, this suggests that the choking occurs at 1/√k and it appears that Shapiro was the first flow model should be called Shapiro’s flow. The author in- one to realize this difference compared to the other models. In reality, the flow is vites others to help in this in- choked somewhere between 1/√k to one for cases that are between Fanno (adi- formation. abatic) and isothermal flow. This fact was evident in industrial applications where the expectation of the choking is at Mach one, but can be explained by choking at a lower Mach number. No experimental evidence, known by the undersigned, was ever produced to verify this finding.

1.3.4 External flow When the flow over an external body is about .8 Mach or more the flow must be considered to be a compressible flow. However at a Mach number above 0.8 (relative of velocity of the body to upstream velocity) a local Mach number (local velocity) can reach M = 1. At that stage, a shock wave occurs which increases the resistance. The Navier-Stokes equations which describe the flow (or even

40This material is very important and someone should find it and make it available to researchers. 41Fanning f based radius is only one quarter of the Darcy f which is based on diameter 14 CHAPTER 1. INTRODUCTION Euler equations) were considered unsolvable during the mid 18xx because of the high complexity. This problem led to two consequences. Theoreticians tried to simplify the equations and arrive at approximate solutions representing specific cases. Examples of such work are Hermann von Helmholtz’s concept of vortex filaments (1858), Lanchester’s concept of circulatory flow (1894), and the Kutta- Joukowski circulation theory of lift (1906). Practitioners like the Wright brothers relied upon experimentation to figure out what theory could not yet tell them. Ludwig Prandtl in 1904 explained the two most important causes of drag by introducing the boundary layer theory. Prandtl’s boundary layer theory allowed various simplifications of the Navier-Stokes equations. Prandtl worked on calculating the effect of induced drag on lift. He introduced the lifting line theory, which was published in 1918-1919 and enabled accurate calculations of induced drag and its effect on lift42. During World War I, Prandtl created his thin–airfoil theory that enabled the calculation of lift for thin, cambered airfoils. He later contributed to the Prandtl- Glauert rule for subsonic airflow that describes the compressibility effects of air at high speeds. Prandtl’s student, Von Karman reduced the equations for supersonic flow into a single equation. After the First World War aviation became important and in the 1920s a push of research focused on what was called the compressibility problem. Air- planes could not yet fly fast, but the propellers (which are also airfoils) did exceed the speed of sound, especially at the propeller tips, thus exhibiting inefficiency. Frank Caldwell and Elisha Fales demonstrated in 1918 that at a critical speed (later renamed the critical Mach number) airfoils suffered dramatic increases in drag and decreases in lift. Later, Briggs and Dryden showed that the problem was related to the shock wave. Meanwhile in Germany, one of Prandtl’s assistants, J. Ack- eret, simplified the shock equations so that they became easy to use. After World War Two, the research had continued and some technical solutions were found. Some of the solutions lead to tedious calculations which lead to the creation of Computational Fluid Dynamics (CFD). Today these methods of perturbations and asymptotic are hardly used in wing calculations43. That is the “dinosaur44” reason that even today some instructors are teaching mostly the perturbations and asymptotic methods in Gas Dynamics classes. More information on external flow can be found in , John D. Anderson’s Book “History of Aerodynamics and Its Impact on Flying Machines,” Cambridge University Press, 1997

42The English call this theory the Lanchester-Prandtl theory. This is because the English Astronomer Frederick Lanchester published the foundation for Prandtl’s theory in his 1907 book Aerodynamics. However, Prandtl claimed that he was not aware of Lanchester’s model when he had begun his work in 1911. This claim seems reasonable in the light that Prandtl was not ware of earlier works when he named erroneously the conditions for the shock wave. See for the full story in the shock section. 43This undersigned is aware of only one case that these methods were really used to calculations of wing. 44It is like teaching using slide ruler in today school. By the way, slide rule is sold for about 7.5$ on the net. Yet, there is no reason to teach it in a regular school. 1.3. HISTORICAL BACKGROUND 15 1.3.5 Filling and Evacuating Gaseous Chambers

It is remarkable that there were so few contributions made in the area of a ﬁlling or evacuation gaseous chamber. The earlier work dealing with this issue was by Giffen, 1940, and was republished by Owczarek, J. A., the model and solution to the nozzle attached to chamber issue in his book “Fundamentals of Gas Dynam- ics.”45. He also extended the model to include the unchoked case. Later several researchers mostly from the University in Illinois extended this work to isothermal nozzle (choked and unchoked).

The simplest model of nozzle, is not sufficient in many cases and a connection by a tube (rather just nozzle or orifice) is more appropriated. Since World War II considerable works have been carried out in this area but with very little progress46. In 1993 the first reasonable models for forced volume were published by the undersigned. Later, that model was extended by several research groups, The analytical solution for forced volume and the “balloon” problem (airbag’s problem) model were published first in this book (version 0.35) in 2005. The classifi- cation of filling or evacuating the chamber as external control and internal control (mostly by pressure) was described in version 0.3 of this book by this author.

1.3.6 Biographies of Major Figures

In this section a short summary of major figures that influenced the field of gas dynamics is present. There are many figures that should be included and a biased selection was required. Much information can be obtained from other resources, such as the Internet. In this section there is no originality and none should be expected.

45International Textbook Co., Scranton, Pennsylvania, 1964. 46In fact, the emergence of the CFD gave the illusion that there are solutions at hand, not realizing that garbage in is garbage out, i.e., the model has to be based on scientiﬁc principles and not detached from reality. As anecdotal story explaining the lack of progress, in die casting conference there was a discussion and presentation on which turbulence model is suitable for a complete still liquid. Other “strange” models can be found in the undersigned’s book “Fundamentals of Die Casting Design. 16 CHAPTER 1. INTRODUCTION

Galileo Galilei

Galileo was born in Pisa, Italy on Febru- ary 15, 1564 to musician Vincenzo Galilei and Giulia degli Ammannati. The oldest of six children, Galileo moved with his family in early 1570 to Florence. Galileo started his studying at the Uni- versity of Pisa in 1581. He then became a professor of mathematics at the Uni- versity of Padua in 1592. During the time after his study, he made numerous discoveries such as that of the pendu- lum clock, (1602). Galileo also proved that objects fell with the same velocity regardless of their size.

Fig. 1.5: Portrait of Galileo Galilei

Galileo had a relationship with Marina Gamba (they never married) who lived and worked in his house in Padua, where she bore him three children. How- ever, this relationship did not last and Marina married Giovanni Bartoluzzi and Galileo’s son, Vincenzio, joined him in Florence (1613).

Galileo invented many mechanical devices such as the pump and the tele- scope (1609). His telescopes helped him make many astronomic observations which proved the Copernican system. Galileo’s observations got him into trouble with the Catholic Church, however, because of his noble ancestry, the church was not harsh with him. Galileo was convicted after publishing his book Dialogue, and he was put under house arrest for the remainder of his life. Galileo died in 1642 in his home outside of Florence. 1.3. HISTORICAL BACKGROUND 17

Ernest Mach (1838-1916) Ernst Mach was born in 1838 in Chrlice (now part of Brno), when Czechia was still a part of the Austro–Hungary empire. Johann, Mach’s father, was a high school teacher who taught Ernst at home until he was 14, when he studied in Kromeriz Gymnasium, before he entered the university of Vi- enna were he studies mathematics, physics and philosophy. He graduated from Vienna in 1860. There Mach wrote his thesis ”On Electrical Discharge and Induction.” Mach was Fig. 1.6: Photo of Ernest Mach interested also in physiology of sen- sory perception. At first he received a professorship position at Graz in mathematics (1864) and was then offered a position as a professor of surgery at the university of Salzburg, but he declined. He then turned to physics, and in 1867 he received a position in the Technical University in Prague47 where he taught experimental physics for the next 28 years. Mach was also a great thinker/philosopher and influenced the theory of relativity dealing with frame of reference. In 1863, Ernest Mach (1836 - 1916) published Die Machanik in which he formalized this argument. Later, Einstein was greatly influenced by it, and in 1918, he named it Mach’s Principle. This was one of the primary sources of inspiration for Einstein’s theory of General Relativity. Mach’s revolutionary experiment demonstrated the existence of the shock wave as shown in Figure 1.7. It is amazing that Mach was able to photograph the phenomenon using the spinning arm technique (no wind tunnel was available at that time and most definitely nothing that Fig. 1.7: The Photo of the bullet in a supersonic flow that could take a photo at super- Mach made. Note it was not taken in a wind tunnel sonic speeds. His experiments required exact timing. He was not able to attach the camera to the arm and utilize the remote control (not existent at that time). Mach’s shadowgraph

47It is interesting to point out that Prague provided us two of the top influential researchers[:] E. Mach and E.R.G. Eckert. 18 CHAPTER 1. INTRODUCTION technique and a related method called Schlieren Photography are still used today. Yet, Mach’s contributions to supersonic flow were not limited to experimental methods alone. Mach understood the basic characteristics of external supersonic flow where the most important variable affecting the flow is the ratio of the speed of the flow48 (U) relative to the speed of sound (c). Mach was the first to note the transition that occurs when the ratio U/c goes from being less than 1 to greater than 1. The name Mach Number (M) was coined by J. Ackeret (Prandtl’s student) in 1932 in honor of Mach.

John William Strutt (Lord Rayleigh) A researcher with a wide interest, started studies in compressible flow mostly from a mathematical approach. At that time there wasn’t the realization that the flow could be choked. It seems that Rayleigh was the first who realized that flow with chemical reactions (heat transfer) can be choked. Lord Rayleigh was a British physicist born near Maldon, Essex, on November 12, 1842. In 1861 he entered Trinity College at Cambridge, where he commenced reading mathematics. His ex- ceptional abilities soon enabled him to overtake his colleagues. He graduated in the Mathe- matical Tripos in 1865 as Senior Wrangler and Smith’s Prizeman. In 1866 he obtained a fellow- ship at Trinity which he held until 1871, the year of his marriage. He served for six years as the Fig. 1.8: Photo of Lord Rayleigh president of the government committee on ex- plosives, and from 1896 to 1919 he acted as Scientific Adviser to Trinity House. He was Lord Lieutenant of Essex from 1892 to 1901. Lord Rayleigh’s first research was mainly mathematical, concerning optics and vibrating systems, but his later work ranged over almost the whole field of physics, covering sound, wave theory, color vision, electrodynamics, electromag- netism, light scattering, flow of liquids, hydrodynamics, density of gases, viscosity, capillarity, elasticity, and photography. Rayleigh’s later work was concentrated on electric and magnetic problems. Rayleigh was considered to be an excellent instructor. His Theory of Sound was published in two volumes during 1877-1878, and his other extensive studies are reported in his Scientific Papers, six volumes issued during 1889-1920. Rayleigh was also a contributer to the Encyclopedia Bri- tannica. He published 446 papers which, reprinted in his collected works, clearly

48Mach dealt with only air, but it is reasonable to assume that he understood that this ratio was applied to other gases. 1.3. HISTORICAL BACKGROUND 19 show his capacity for understanding everything just a little more deeply than anyone else. He intervened in debates of the House of Lords only on rare occasions, never allowing politics to interfere with science. Lord Rayleigh, a Chancellor of Cambridge University, was a Justice of the Peace and the recipient of honorary science and law degrees. He was a Fellow of the Royal Society (1873) and served as Secretary from 1885 to 1896, and as President from 1905 to 1908. He received the Nobel Prize in 1904. Lord Rayleigh died on June 30, 1919, at Witham, Essex.

In 1871 he married Evelyn, sister of the future prime minister, the Earl of Balfour (of the famous Balfour declaration of the Jewish state). They had three sons, the eldest of whom was to become a professor of physics at the Imperial College of Science and Technology, London.

As a successor to James Clerk Maxwell, he was head of the Cavendish Laboratory at Cambridge from 1879-1884, and in 1887 became Professor of Nat- ural Philosophy at the Royal Institute of Great Britain. Rayleigh died on June 30, 1919 at Witham, Essex.

William John Macquorn Rankine

William John Macquorn Rankine (July 2, 1820 - December 24, 1872) was a Scottish engineer and physicist. He was a founding contributor to the science of thermodynamics (Rankine Cycle). Rankine developed a theory of the steam engine. His steam engine manuals were used for many decades. Rankine was well rounded interested beside the energy ﬁeld he was also interested in civil engineering, strength of materials, and naval engineering in which he was involved in applying scientiﬁc principles to building ships. Rankine was born in Edinburgh to British Army lieutenant David Rankine and Barbara Fig. 1.9: Portrait of Rankine Grahame, Rankine. Rankine never married, and his only brother and parents died before him. 20 CHAPTER 1. INTRODUCTION

Gino Girolamo Fanno

Fanno a Jewish Engineer was born on November 18, 1888. He studied in a technical institute in Venice and graduated with very high grades as a mechanical engineer. Fanno was not as lucky as his brother, who was able to get into academia. Faced with anti–semitism, Fanno left Italy for Zurich, Switzerland in 1900 to attend graduate school for his master’s degree. In this new place he was able to pose as a Ro- man Catholic, even though for short time he went to live in a Jewish home, Isaak Baruch Weil’s family. As were many Jews at that time, Fanno was fluent in several lan- guages including Italian, English, German, Fig. 1.10: The photo of Gino Fanno ap- and French. He likely had a good knowl- proximately in 1950 edge of Yiddish and possibly some Hebrew. Consequently, he did not have a problem studying in a different language. In July 1904 he received his diploma (master). When one of Professor Stodola’s assistants attended military service this temporary position was offered to Fanno. “Why didn’t a talented guy like Fanno keep or obtain a position in academia after he published his model?” The answer is tied to the fact that somehow rumors about his roots began to surface. Additionally, the fact that his model was not a “smash- ing49success” did not help. Later Fanno had to go back to Italy to find a job in industry. Fanno turned out to be a good engineer and he later obtained a management position. He married, and like his brother, Marco, was childless. He obtained a Ph.D. from Regian Istituto Superiore d’Ingegneria di Genova. However, on February 1939 Fanno was degraded (denounced) and he lost his Ph.D. (is this the first case in history) because his of his Jewish nationality50. During the War (WWII), he had to be under house arrest to avoid being sent to the “vacation camps.” To further camouflage himself, Fanno converted to Catholicism. Apparently, Fanno had a cache of old Italian currency (which was apparently still highly acceptable) which helped him and his wife survive the war. After the war, Fanno was only able to work in agriculture and agricultural engineering. Fanno passed way in 1960 without world recognition for his model. Fanno’s older brother, mentioned earlier Marco Fanno is a famous economist who later developed fundamentals of the supply and demand theory.

49Missing data about friction factor 50In some places, the ridicules claims that Jews persecuted only because their religion. Clearly, Fanno was not part of the Jewish religion (see his picture) only his nationality was Jewish. 1.3. HISTORICAL BACKGROUND 21

Ludwig Prandtl Perhaps Prandtl’s greatest achievement was his ability to produce so many great scientists. It is mind boggling to look at the long list of those who were his students and colleagues. There is no one who educated as many great scientists as Prandtl. Prandtl changed the field of fluid mechanics and is called the modern father of fluid mechanics because of his introduction of boundary layer, turbulence mixing theories etc. Ludwig Prandtl was born in Freising, Bavaria, in 1874. His father was a professor of engineering and his mother suffered from a lengthy illness. As a result, the young Ludwig spent more time with his father which made him interested in his father’s physics and machinery books. This upbringing fostered Fig. 1.11: Photo of Prandtl the young Prandtl’s interest in science and experimentation. Prandtl started his studies at the age of 20 in Munich, Germany and he graduated at the age of 26 with a Ph.D. Interestingly, his Ph.D. was focused on solid mechanics. His interest changed when, in his first job, he was required to design factory equipment that involved problems related to the field of fluid mechanics (a suction device). Later he sought and found a job as a professor of mechanics at a technical school in Hannover, Germany (1901). During this time Prandtl developed his boundary layer theory and studied supersonic fluid flows through nozzles. In 1904, he presented the revolutionary paper “Flussigkeitsbewe- gung Bei Sehr Kleiner Reibung” (Fluid Flow in Very Little Friction), the paper which describes his boundary layer theory. His 1904 paper raised Prandtl’s prestige. He became the director of the Institute for Technical Physics at the University of Gottingen.¨ He developed the Prandtl-Glauert rule for subsonic airflow. Prandtl, with his student Theodor Meyer, developed the first theory for calculating the properties of shock and expansion waves in supersonic flow in 1908 (two chapters in this book). As a byproduct they produced the theory for oblique shock. In 1925 Prandtl became the director of the Kaiser Wilhelm Institute for Flow Investigation at Gottingen.¨ By the 1930s, he was known worldwide as the leader in the science of fluid dynamics. Prandtl also contributed to research in many areas, such as meteorology and structural mechanics. Ludwig Prandtl worked at Gottingen¨ until his death on August 15, 1953. His work and achievements in fluid dynamics resulted in equations that simplified 22 CHAPTER 1. INTRODUCTION understanding, and many are still used today. Therefore many referred to him as the father of modern fluid mechanics. Ludwig Prandtl died in Gottingen,¨ Germany on August 15th 1953. Prandtl’s other contributions include: the introduction of the Prandtl number in fluid mechanics, airfoils and wing theory (including theories of aerodynamic in- terference, wing-fuselage, wing-propeller, biplane, etc); fundamental studies in the wind tunnel, high speed flow (correction formula for subsonic compressible flows), theory of turbulence. His name is linked to the following:

• Prandtl number (heat transfer problems)

• Prandtl-Glauert compressibility correction

• Prandtl’s boundary layer equation

• Prandtl’s lifting line theory

• Prandtl’s law of friction for smooth pipes

• Prandtl-Meyer expansion fans (supersonic ﬂow)

• Prandtl’s Mixing Length Concept (theory of turbulence)

E.R.G. Eckert

Eckert was born in 1904 in Prague, where he studied at the German Insti- tute of Technology. During World War II, he developed methods for jet engine turbine blade cooling at a research laboratory in Prague. He emigrated to the United States after the war, and served as a consultant to the U.S. Air Force and the National Advisory Committee for Aeronautics before coming to Min- nesota. Eckert developed the under- Fig. 1.12: The photo of Ernst Rudolf George standing of heat dissipation in relation to Eckert with the author’s family kinetic energy, especially in compressible ﬂow. Hence, the dimensionless group has been designated as the Eckert number, which is associated with the Mach number. Schlichting suggested this dimensionless group in honor of Eckert. In addition to being named to the National Academy of Engineering in 1970, He authored more than 500 articles and received several medals for his contributions to science. His book ”Introduction to the Transfer of Heat and Mass,” published in 1937, is still considered a fundamental text in the ﬁeld. 1.3. HISTORICAL BACKGROUND 23 Eckert was an excellent mentor to many researchers (including this author), and he had a reputation for being warm and kindly. He was also a leading Figure in bringing together engineering in the East and West during the Cold War years.

Ascher Shapiro MIT Professor Ascher Shapiro51, the Eckert equivalent for the compressible flow, was instrumental in using his two volume book “The Dynamics of Thermodynamics of the Compressible Fluid Flow,” to transform the gas dynamics field to a coherent text material for engineers. Furthermore, Shapiro’s knowledge of fluid mechanics enabled him to “sew” the missing parts of the Fanno line with Moody’s diagram to create the most useful model in compressible flow. While Shapiro viewed gas dynamics mostly through aeronautic eyes, The undersigned believes that Shapiro was the first one to propose an isothermal flow model that is not part of the aeronautic field. Therefore it is proposed to call this model Shapiro’s Flow. In his first 25 years Shapiro focused primarily on power production, high- speed flight, turbomachinery and propulsion by jet engines and rockets. Unfor- tunately for the field of Gas Dynamics, Shapiro moved to the field of biomedical engineering where he was able to pioneer new work. Shapiro was instrumental in the treatment of blood clots, asthma, emphysema and glaucoma. Shapiro grew up in New York City and received his S.B. in 1938 and the Sc.D. (It is M.I.T.’s equivalent of a Ph.D. degree) in 1946 in mechanical engineering from MIT. He was assistant professor in 1943, three years before receiving his Sc.D. In 1965 he become the head of the Department of Mechanical Engineering until 1974. Shapiro spent most of his active years at MIT. Ascher Shapiro passed way in November 2004.

51Parts taken from Sasha Brown, MIT 24 CHAPTER 1. INTRODUCTION CHAPTER 2

Fundamentals of Basic Fluid Mechanics

2.1 Introduction This chapter is a review of the fundamentals that the student is expected to know. The basic principles are related to the basic conservation principle. Several terms will be reviewed such as stream lines. In addition the basic Bernoulli’s equation will be derived for incompressible flow and later for compressible flow. Several application of the fluid mechanics will demonstrated. This material is not covered in the history chapter.

2.2 Fluid Properties 2.3 Control Volume 2.4 Reynold’s Transport Theorem For simplification the discussion will be focused on one dimensional control volume and it will be generalzed later. The flow through a stream tube is assumed to be one-dimensional so that there isn’t any flow except at the tube opening. At the initial time the mass that was in the tube was m0. The mass after a very short time of dt is dm. For simplicity, it is assumed the control volume is a fixed boundary. The flow on the right through the opening and on the left is assumed to enter the stream tube while the flow is assumed to leave the stream tube.

25 26 CHAPTER 2. FUNDAMENTALS OF BASIC FLUID MECHANICS Supposed that the ﬂuid has a property η

dN N (t + ∆t) N (t ) s = lim s 0 − s 0 (2.1) dt ∆t 0 ∆t → CHAPTER 3

Speed of Sound

3.1 Motivation In traditional compressible flow classes there is very little discussion about the speed of sound outside the ideal gas. The author thinks that this approach has many shortcomings. In a recent consultation an engineer1 design a industrial system that contains converting diverging nozzle with filter to remove small particles from air. The engineer was well aware of the calculation of the nozzle. Thus, the engineer was able to predict that was a chocking point. Yet, the engineer was not ware of the effect of particles on the speed of sound. Hence, the actual flow rate was only half of his prediction. As it will shown in this chapter, the particles can, in some situations, reduces the speed of sound by almost as half. With the “new” knowledge from the consultation the calculations were within the range of acceptable results. The above situation is not unique in the industry. It should be expected that engineers know how to manage this situation of non pure substances (like clean air). The fact that the engineer knows about the chocking is great but it is not enough for today’s sophisticated industry2. In this chapter an introductory discussion is given about different situations which can appear the industry in regards to speed of sound.

3.2 Introduction

1Aerospace engineer that alumni of University of Minnesota, Aerospace Department. 2Pardon, but a joke is must in this situation. A cat is pursuing a mouse and the mouse escape and hide in the hole. Suddenly, the mouse hear a barking dog and a cat yelling. The mouse go out to investigate, and cat is catching the mouse. The mouse ask the cat I thought I hear a dog. The cat reply, yes you right. My teacher was right, one language is not enough today.

27 28 CHAPTER 3. SPEED OF SOUND The people had recognized for several hundred years that sound is sound wave dU velocity=dU a variation of pressure. The ears c sense the variations by frequency P+dP P and magnitude which are trans- ρ+dρ ρ ferred to the brain which translates to voice. Thus, it raises the question: what is the speed of the Fig. 3.1: A very slow moving piston in a still gas small disturbance travel in a “quiet” medium. This velocity is referred to as the speed of sound. To answer this question consider a piston moving from the left to the right at a relatively small velocity (see Figure 3.1). The information that the piston is moving passes thorough a single “pressure pulse.” It is assumed that if the velocity of the piston is inﬁnitesimally small, the pulse will be inﬁnitesimally small. Thus, the pressure and density can be assumed to be continuous. In the control volume it is convenient to look at a control vol- Control volume around ume which is attached to a pressure the sound wave pulse. Applying the mass balance c-dU c yields P+dP P ρ+dρ ρ ρc = (ρ + dρ)(c dU) (3.1) − or when the higher term dUdρ is Fig. 3.2: Stationary sound wave and gas moves neglected yields relative to the pulse cdρ ρdU = cdρ = dU = (3.2) ⇒ ρ

From the energy equation (Bernoulli’s equation), assuming isentropic ﬂow and neglecting the gravity results

(c dU)2 c2 dP − − + = 0 (3.3) 2 ρ

neglecting second term (dU 2) yield dP cdU + = 0 (3.4) − ρ

Substituting the expression for dU from equation (3.2) into equation (3.4) yields

dρ dP dP c2 = = c2 = (3.5) ρ ρ ⇒ dρ An expression is needed to represent the right hand side of equation (3.5). For an ideal gas, P is a function of two independent variables. Here, it is considered 3.3. SPEED OF SOUND IN IDEAL AND PERFECT GASES 29 that P = P (ρ, s) where s is the entropy. The full differential of the pressure can be expressed as follows:

∂P ∂P dP = dρ + ds (3.6) ∂ρ ∂s s ρ

In the derivations for the speed of sound it was assumed that the ﬂow is isentropic, therefore it can be written

dP ∂P = (3.7) dρ ∂ρ s

Note that the equation (3.5) can be obtained by utilizing the momentum equation instead of the energy equation.

Example 3.1: Demonstrate that equation (3.5) can be derived from the momentum equation.

SOLUTION The momentum equation written for the control volume shown in Figure (3.2) is

F cs U(ρUdA) R (P +PdP ) P = (ρ + dρ)(c dU)2 ρc2 (3.8) − − − z }| { z }| { Neglecting all the relative small terms results in

: 0 : 0 dP = (ρ + dρ) c2 2cdU +∼ dU 2 ∼ ρc2 (3.9) − ! −

dP = c2dρ (3.10)

This yields the same equation as (3.5).

3.3 Speed of sound in ideal and perfect gases The speed of sound can be obtained easily for the equation of state for an ideal gas (also perfect gas as a sub set) because of a simple mathematical expression. The pressure for an ideal gas can be expressed as a simple function of density, ρ, and a function “molecular structure” or ratio of speciﬁc heats, k namely

P = constant ρk (3.11) × 30 CHAPTER 3. SPEED OF SOUND and hence P k dP k 1 constant ρ c = = k constant ρ − = k × s dρ × × × z }|ρ { P = k (3.12) × ρ Remember that P/ρ is deﬁned for an ideal gas as RT , and equation (3.12) can be written as c = √kRT (3.13)

Example 3.2: Calculate the speed of sound in water vapor at 20[bar] and 350◦C, (a) utilizes the steam table (b) assuming ideal gas.

SOLUTION The solution can be estimated by using the data from steam table3

∆P c = (3.14) ∆ρ s s=constant

kJ kg At 20[bar] and 350◦C: s = 6.9563 K kg ρ = 6.61376 m3 h kJ i hkg i At 18[bar] and 350◦C: s = 7.0100 K kg ρ = 6.46956 m3 h kJ i h kg i At 18[bar] and 300◦C: s = 6.8226 K kg ρ = 7.13216 m3 h i h i After interpretation of the temperature: kJ kg At 18[bar] and 335.7◦C: s 6.9563 ρ 6.94199 3 ∼ K kg ∼ m and substituting into the equation yieldsh i h i 200000 m c = = 780.5 (3.15) 0.32823 sec r h i for ideal gas assumption (data taken from Van Wylen and Sontag, Classical Ther- modynamics, table A 8.) m c = √kRT 1.327 461 (350 + 273) 771.5 ∼ × × ∼ sec p h i Note that a better approximation can be done with a steam table, and it will be part of the future program (potto–GDC).

3This data is taken form Van Wylen and Sontag “Fundamentals of Classical Thermodynamics” 2nd edition 3.4. SPEED OF SOUND IN REAL GAS 31 Example 3.3: The temperature in the atmosphere can be assumed to be a linear function of the height for some distances. What is the time it take for sound to travel from point “A” to point “B” under this assumption.?

SOLUTION The temperature is denoted at “A” as TA and temperature in “B” is TB. The distance between “A” and “B” is denoted as h. x T = (T T ) + T B − A h A Where the distance x is the variable distance. It should be noted that velocity is provided as a function of the distance and not the time (another reverse problem). For an inﬁnitesimal time dt is equal to dx dx dt = = kRT (x) (TB TA)x kRTA − + 1 TAh p r integration of the above equation yields

3 2hT T 2 t = A B 1 (3.16) 3√kRT (T T ) TA − A B − A ! For assumption of constant temperature the time is h t = (3.17) √kRT¯ Hence the correction factor

3 t T 2 T T 2 corrected = A A B 1 (3.18) t T¯ 3 (T T ) T − r B − A A ! This correction factor approaches one when T T . B −→ A 3.4 Speed of Sound in Real Gas The ideal gas model can be improved by introducing the compressibility factor. The compressibility factor represents the deviation from the ideal gas. Thus, a real gas equation can be expressed in many cases as

P = zρRT (3.19)

The speed of sound of any gas is provided by equation (3.7). To obtain the expression for a gas that obeys the law expressed by (3.19) some mathematical expressions are needed. Recalling from thermodynamics, the Gibbs function (3.20) 32 CHAPTER 3. SPEED OF SOUND

Fig. 3.3: The Compressibility Chart is used to obtain dP T ds = dh (3.20) − ρ The definition of pressure specific heat for a pure substance is ∂h ∂s C = = T (3.21) p ∂T ∂T P P The definition of volumetric specific heat for a pure substance is ∂u ∂s C = = T (3.22) v ∂T ∂T ρ ρ From thermodynamics, it can be shown 4 ∂v dh = C dT + v T (3.23) p − ∂T P 4See Van Wylen p. 372 SI version, perhaps to insert the discussion here. 3.4. SPEED OF SOUND IN REAL GAS 33 The specific volumetric is the inverse of the density as v = zRT/P and thus 1 zRT > ∂v ∂ RT ∂z zR ∂T = P = + (3.24) ∂T ∂T P ∂T P ∂T P !P P P Substituting the equation (3.24) into equation (3.23) results

v v z T RT ∂z zR dh = CpdT + v T  +  dP (3.25) − z}|{P ∂T P z}|{P             Simplifying equation (3.25) to became T v ∂z T ∂z dP dh = C dT dP = C dT (3.26) p − z ∂T p − z ∂T ρ P P Utilizing Gibbs equation (3.20)

dh T ∂z dP dP dP T ∂z T ds = C dT = C dT + 1 p − z ∂T ρ − ρ p − ρ z ∂T z }| P { P zRT dP P T ∂z =C dT + 1 (3.27) p − P ρ z ∂T z}|{ P Letting ds = 0 for isentropic process results in dT dP R ∂z = z + T (3.28) T P C ∂T p P Equation (3.28) can be integrated by parts. However, it is more convenient to express dT/T in terms of Cv and dρ/ρ as follows

dT dρ R ∂z = z + T (3.29) T ρ C ∂T v " ρ# Equating the right hand side of equations (3.28) and (3.29) results in

dρ R ∂z dP R ∂z z + T = z + T (3.30) ρ C ∂T P C ∂T v " ρ# p P Rearranging equation (3.30) yields

dρ dP C z + T ∂z = v ∂T P (3.31) ρ P C ∂z p " z + T ∂T ρ # 34 CHAPTER 3. SPEED OF SOUND If the terms in the braces are constant in the range under interest in this study, equation (3.31) can be integrated. For short hand writing convenience, n is deﬁned as

k ∂z C z + T ∂T n = p ρ (3.32) C ∂z z}|{v z + T ∂T P ! Note that n approaches k when z 1 and when z is constant. The integration of → equation (3.31) yields

ρ n P 1 = 1 (3.33) ρ P 2 2 Equation (3.33) is similar to equation (3.11). What is different in these derivations is that a relationship between coefﬁcient n and k was established. This relationship (3.33) isn’t new, and in–fact any thermodynamics book shows this relationship. But the deﬁnition of n in equation (3.32) provides a tool to estimate n. Now, the speed of sound for a real gas can be obtained in the same manner as for an ideal gas. dP = nzRT (3.34) dρ

Example 3.4: Calculate the speed of sound of air at 30◦C and atmospheric pressure 1[bar]. ∼ The speciﬁc heat for air is k = 1.407, n = 1.403, and z = 0.995. Make the calculation based on the ideal gas model and compare these calculations to real gas model (compressibility factor). Assume that R = 287[j/kg/K].

SOLUTION According to the ideal gas model the speed of sound should be

c = √kRT = √1.407 287 300 348.1[m/sec] × × ∼ For the real gas ﬁrst coefﬁcient n = 1.403 has

c = √znRT = √1.403 0.995times287 300 = 346.7[m/sec] × ×

The correction factor for air under normal conditions (atmospheric conditions or even increased pressure) is minimal on the speed of sound. However, a change in temperature can have a dramatical change in the speed of sound. For example, at relative moderate pressure but low temperature common in atmosphere, the compressibility factor, z = 0.3 and n 1 which means that speed of sound is only ∼ 0.3 1.4 factor (0.5) to calculated by ideal gas model. q 3.5. SPEED OF SOUND IN ALMOST INCOMPRESSIBLE LIQUID 35 3.5 Speed of Sound in Almost Incompressible Liquid Even liquid normally is assumed to be incompressible in reality has a small and important compressible aspect. The ratio of the change in the fractional volume to pressure or compression is referred to as the bulk modulus of the material. For example, the average bulk modulus for water is 2.2 109 N/m2. At a depth of about × 4,000 meters, the pressure is about 4 107 N/m2. The fractional volume change × is only about 1.8% even under this pressure nevertheless it is a change. The compressibility of the substance is the reciprocal of the bulk modulus. The amount of compression of almost all liquids is seen to be very small as given in Table (3.5). The mathematical deﬁnition of bulk modulus as following dP B = ρ (3.35) dρ

In physical terms can be written as

elastic property B c = = (3.36) sinertial property s ρ

For example for water

2.2 109N/m2 c = × = 1493m/s s 1000kg/m3

This agrees well with the measured speed of sound in water, 1482 m/s at 20◦C. Many researchers have looked at this velocity, and for purposes of comparison it is given in Table (3.5)

Remark reference Value [m/sec] Fresh Water (20 ◦C) Cutnell, John D. & Kenneth W. 1492 Johnson. Physics. New York: Wiley, 1997: 468. Distilled Water at (25 ◦C) The World Book Encyclopedia. 1496 Chicago: World Book, 1999. 601 Water distilled Handbook of Chemistry and 1494 Physics. Ohio: Chemical Rubber Co., 1967-1968: E37

Table 3.1: Water speed of sound from different sources

The effect of impurity and temperature is relatively large, as can be observed from the equation (3.37). For example, with an increase of 34 degrees from 0◦C there is an increase in the velocity from about 1430 m/sec to about 1546 [m/sec]. According 36 CHAPTER 3. SPEED OF SOUND to Wilson5, the speed of sound in sea water depends on temperature, salinity, and hydrostatic pressure. Wilson’s empirical formula appears as follows:

c(S, T, P ) = c0 + cT + cS + cP + cST P , (3.37)

where c0 = 1449.14 is about clean/pure water, cT is a function temperature, and cS is a function salinity, cP is a function pressure, and cST P is a correction factor between coupling of the different parameters.

material reference Value [m/sec] Glycerol 1904 Sea water 25 ◦C 1533 Mercury 1450 Kerosene 1324 Methyl alcohol 1143 Carbon tetrachloride 926

Table 3.2: Liquids speed of sound, after Aldred, John, Manual of Sound Recording, London: Fountain Press, 1972

In summary, the speed of sound in liquids is about 3 to 5 relative to the speed of sound in gases.

3.6 Speed of Sound in Solids The situation with solids is considerably more complicated, with different speeds in different directions, in different kinds of geometries, and differences between transverse and longitudinal waves. Nevertheless, the speed of sound in solids is larger than in liquids and deﬁnitely larger than in gases. Young’s Modulus for a representative value for the bulk modulus for steel is 160 109 N /m2. Speed of sound in solid of steel, using a general tabulated value for the bulk modulus, gives a sound speed for structural steel of

E 160 109N/m2 c = = × = 4512m/s (3.38) s ρ s 7860Kg/m3

Compared to one tabulated value the example values for stainless steel lays between the speed for longitudinal and transverse waves.

5J. Acoust. Soc. Amer., 1960, vol.32, N 10, p. 1357. Wilson’s formula is accepted by the National Oceanographic Data Center (NODC) USA for computer processing of hydrological information. 3.7. SOUND SPEED IN TWO PHASE MEDIUM 37 material reference Value [m/sec] Diamond 12000 Pyrex glass 5640 Steel longitudinal wave 5790 Steel transverse shear 3100 Steel longitudinal wave (extensional 5000 wave) Iron 5130 Aluminum 5100 Brass 4700 Copper 3560 Gold 3240 Lucite 2680 Lead 1322 Rubber 1600

Table 3.3: Solids speed of sound, after Aldred, John, Manual of Sound Recording, Lon- don:Fountain Press, 1972

3.7 Sound Speed in Two Phase Medium The gas flow in many industrial situations contains other particles. In actuality, there could be more than one speed of sound for two phase flow. Indeed there is double chocking phenomenon in two phase flow. However, for homogeneous and under certain condition a single velocity can be considered. There can be several models that approached this problem. For simplicity, it assumed that two materials are homogeneously mixed. Topic for none homogeneous mixing are beyond the scope of this book. It further assumed that no heat and mass transfer occurs between the particles. In that case, three extreme cases suggest themselves: the flow is mostly gas with drops of the other phase (liquid or solid), about equal parts of gas and the liquid phase, and liquid with some bubbles. The first case is analyzed. The equation of state for the gas can be written as

Pa = ρaRTa (3.39) The average density can be expressed as 1 ξ 1 ξ = + − (3.40) ρm ρa ρb

m˙ b where ξ = m˙ is the mass ratio of the materials. For small value of ξ equation (3.40) can be approximated as ρ = 1 + m (3.41) ρa 38 CHAPTER 3. SPEED OF SOUND

where m = m˙ b is mass flow rate per gas flow rate. m˙ a The gas density can be replaced by equation (3.39) and substituted into equation (3.41) P R = T (3.42) ρ 1 + m A approximation of addition droplets of liquid or dust (solid) results in reduction of R and yet approximate equation similar to ideal gas was obtained. It must noticed that m = constant. If the droplets (or the solid particles) can be assumed to have the same velocity as the gas with no heat transfer or fiction between the particles isentropic relation can be assumed as P k = constant (3.43) ρa Assuming that partial pressure of the particles is constant and applying the second law for the mixture yields

droplets gas

dT dT dP (Cp + mC)dT dP 0 = mC + Cp R = R (3.44) z }|T{ z T }|− P{ T − P Therefore, the mixture isentropic relationship can be expressed as

γ 1 P −γ = constant (3.45) T where γ 1 R − = (3.46) γ Cp + mC Recalling that R = C C reduces equation (3.46) into p − v C + mC γ = p (3.47) Cv + mC In a way the definition of γ was so chosen that effective specific pressure heat Cp+mC Cv+mC and effective specific volumetric heat are 1+m and 1+m respectively. The correction factors for the specific heat is not linear. Since the equations are the same as before hence the familiar equation for speed of sound can be applied as

c = γRmixT (3.48)

Insert example with small It can be noticed that Rmix and γ arep smaller than similar variables in a pure steel particles with air up to 20% gas. Hence, this analysis results in lower speed of sound compared to pure gas. Generally, the velocity of mixtures with large gas component is smaller of the pure gas. For example, the velocity of sound in slightly wed steam can be about one At this stage the other models third of the pure steam speed of sound. for two phase are left for next version (0.6). 3.7. SOUND SPEED IN TWO PHASE MEDIUM 39 Meta For a mixture of two phases, speed of sound can be expressed as

∂P ∂P [f(X)] c2 = = (3.49) ∂ρ ∂ρ

where X is deﬁned as s s (P ) X = − f B (3.50) sfg(PB) Meta End 40 CHAPTER 3. SPEED OF SOUND CHAPTER 4

Isentropic Flow

In this chapter a discussion on a steady state flow through a smooth and continuous area flow rate is presented. A discussion about the flow through a converging–diverging nozzle is also part of this chapter. The isentropic flow P = P models are important because of two B 0 main reasons: One, it provides the infor- P P0 Subsonic mation about the trends and important M < 1 parameters. Two, the correction factors can be introduced later to account for Supersonic deviations from the ideal state. M > 1 distance, x

4.1 Stagnation State for Fig. 4.1: Flow of a compressible substance Ideal Gas Model (gas) through a converging– diverging nozzle. 4.1.1 General Relationship

It is assumed that the flow is one– dimensional. Figure (4.1) describes a gas flow through a converging–diverging nozzle.t has been found that a theoretical state known as the stagnation state is very useful in simplifying the solution and treatment of the flow. he stagnation state is a theoretical state in which the flow is brought into a complete motionless condition in isentropic process without other forces (e.g. gravity force). Several properties that can be represented by this theoretical process which include temperature, pressure, and density et cetera and denoted by the subscript “0.”

41 42 CHAPTER 4. ISENTROPIC FLOW First, the stagnation temperature is calculated. The energy conservation can be written as U 2 h + = h (4.1) 2 0

Perfect gas is an ideal gas with a constant heat capacity, Cp. For perfect gas equation (4.1) is simpliﬁed into

U 2 C T + = C T (4.2) p 2 p 0

Again it is common to denote T0 as the stagnation temperature. Recalling from thermodynamic the relationship for perfect gas

R = C C (4.3) p − v and denoting k C C then the thermodynamics relationship obtains the form ≡ p ÷ v kR C = (4.4) p k 1 − and where R is a speciﬁc constant. Dividing equation (4.2) by (CpT ) yields

2 U T0 1 + = (4.5) 2CpT T

Now, substituting c2 = kRT or T = c2/kR equation (4.5) changes into

2 kRU T0 1 + 2 = (4.6) 2Cpc T

By utilizing the deﬁnition of k by equation (4.4) and inserting it into equation (4.6) yields

k 1 U 2 T 1 + − = 0 (4.7) 2 c2 T

It very useful to convert equation (4.6) into a dimensionless form and denote Mach number as the ratio of velocity to speed of sound as U M (4.8) ≡ c nserting the deﬁnition of Mach number (4.8) into equation (4.7) reads T k 1 0 = 1 + − M 2 (4.9) T 2

4.1. STAGNATION STATE FOR IDEAL GAS MODEL 43

The usefulness of Mach num- A B ber and equation (4.9) can be demon- T0 T0 strated by this following simple example. P0 P0 ρ0 velocity ρ0 In this example a gas flows through a tube (see Figure 4.2) of any shape can be expressed as a function of only the stagnation temperature as opposed to Fig. 4.2: Perfect gas flows through a tube the function of the temperatures and velocities. The definition of the stagnation state provides the advantage of compact writing. For example, writing the energy equation for the tube shown in Figure (4.2) can be reduced to

Q˙ = C (T T )m˙ (4.10) p 0B − 0A

The ratio of stagnation pressure to the static pressure can be expressed as the function of the temperature ratio because of the isentropic relationship as

k k k 1 k 1 P T − k 1 − 0 = 0 = 1 + − M 2 (4.11) P T 2 In the same manner the relationship for the density ratio is

1 1 k 1 k 1 ρ T − k 1 − 0 = 0 = 1 + − M 2 (4.12) ρ T 2 A new useful deﬁnition is introduced for the case when M = 1 and denoted by superscript “ .” The special case of ratio of the star values to stagnation values are ∗ dependent only on the heat ratio as the following:

2 T ∗ c∗ 2 = 2 = (4.13) T0 c0 k + 1

k k 1 P ∗ 2 − = (4.14) P k + 1 0

1 k 1 ρ∗ 2 − = (4.15) ρ k + 1 0 44 CHAPTER 4. ISENTROPIC FLOW Static Properties As A Function of Mach Number

0.9

0.8 P/P0 ρ/ρ 0.7 0 T/T0 0.6

0.5

0.4

0.3

0.2

0.1

0 0 1 2 3 4 5 6 7 8 9 Mon Jun 5 17:39:34 2006 Mach number

Fig. 4.3: The stagnation properties as a function of the Mach number, k = 1.4

4.1.2 Relationships for Small Mach Number Even with today’s computers a simpliﬁed method can reduce the tedious work involved in computational work. In particular, the trends can be examined with analytical methods. It further will be used in the book to examine trends in derived models. It can be noticed that the Mach number involved in the above equations is in a square power. Hence, if an acceptable error is of about %1 then M < 0.1 provides the desired range. Further, if a higher power is used, much smaller error results. First it can be noticed that the ratio of temperature to stagnation temperature, T is provided in power series. Expanding of the equations according to the T0 binomial expansion of

n(n 1)x2 n(n 1)(n 2)x3 (1 + x)n = 1 + nx + − + − − + (4.16) 2! 3! · · · will resutls in the same fashion P (k 1)M 2 kM 4 2(2 k)M 6 0 = 1 + − + + − (4.17) P 4 8 48 · · · 4.2. ISENTROPIC CONVERGING-DIVERGING FLOW IN CROSS SECTION 45

ρ (k 1)M 2 kM 4 2(2 k)M 6 0 = 1 + − + + − (4.18) ρ 4 8 48 · · ·

The pressure difference normalized by the velocity (kinetic energy) as correction factor is

compressibility correction P P M 2 (2 k)M 4 0 = 1 + + + 1 − 2 − (4.19) 2 ρU z 4 }|24 · ·{· From the above equation, it can be observed that the correction factor approaches zero when M 0 and then equation (4.19) approaches the standard equation −→ for incompressible ﬂow. The deﬁnition of the star Mach is ratio of the velocity and star speed of soundat M = 1.

U k + 1 k 1 2 M ∗ = = M 1 − M + (4.20) c 2 − 4 · · · ∗ r

P P kM 2 M 2 0 − = 1 + + (4.21) P 2 4 · · ·

ρ ρ M 2 kM 2 0 − = 1 + (4.22) ρ 2 − 4 · · · The normalized mass rate becomes

m˙ kP 2M 2 k 1 = 0 1 + − M 2 + (4.23) A s RT0 4 · · · The ratio of the area to star area is

k+1 2(k 1) A 2 − 1 k + 1 (3 k)(k + 1) = + M + − M 3 + (4.24) A k + 1 M 4 32 · · · ∗

4.2 Isentropic Converging-Diverging Flow in Cross Sec- tion 46 CHAPTER 4. ISENTROPIC FLOW The important sub case in this chapter is the flow in a converging–diverging T T+dT nozzle. The control volume is shown ρ ρ+dρ P P+dP in Figure (4.4). There are two mod- U U+dU els that assume variable area flow: First is isentropic and adiabatic model. Second is isentropic and isothermal Fig. 4.4: Control volume inside a converging- model. Clearly, the stagnation tem- diverging nozzle. perature, T0, is constant through the adiabatic flow because there isn’t heat transfer. Therefore, the stagnation pressure is also constant through the flow because the flow isentropic. Conversely, in mathematical terms, equation (4.9) and equation (4.11) are the same. If the right hand side is constant for one variable, it is constant for the other. In the same argument, the stagnation density is constant through the flow. Thus, knowing the Mach number or the temperature will provide all that is needed to find the other properties. The only properties that need to be connected are the cross section area and the Mach number. Examination of the relation between properties can then be carried out.

4.2.1 The Properties in the Adiabatic Nozzle hen there is no external work and heat transfer, the energy equation, reads

dh + UdU = 0 (4.25)

Differentiation of continuity equation, ρAU = m˙ = constant, and dividing by the continuity equation reads

dρ dA dU + + = 0 (4.26) ρ A U

The thermodynamic relationship between the properties can be expressed as

dP T ds = dh (4.27) − ρ

For isentropic process ds 0 and combining equations (4.25) with (4.27) yields ≡ dP + UdU = 0 (4.28) ρ

Differentiation of the equation state (perfect gas), P = ρRT , and dividing the results by the equation of state (ρRT ) yields

dP dρ dT = + (4.29) P ρ T 4.2. ISENTROPIC CONVERGING-DIVERGING FLOW IN CROSS SECTION 47 Obtaining an expression for dU/U from the mass balance equation (4.26) and using it in equation (4.28) reads

dU U dP dA dρ U 2 + = 0 (4.30) ρ − A ρ z }| { Rearranging equation (4.30) so that the density, dρ, can be replaced by the static pressure, dP/ρ yields

1 c2 dP dA dρ dP dA dρ dP  = U 2 + = U 2 + (4.31) ρ A ρ dP A dP ρ  z}|{          Recalling that dP/dρ = c2 and substitute the speed of sound into equation (4.31) to obtain

dP U 2 dA 1 = U 2 (4.32) ρ − c A " # Or in a dimensionless form dP dA 1 M 2 = U 2 (4.33) ρ − A Equation (4.33) is a differential equation for the pressure as a function of the cross section area. It is convenient to rearrange equation (4.33) to obtain a variables separation form of

ρU 2 dA dP = (4.34) A 1 M 2 −

The pressure Mach number relationship

Before going further in the mathematical derivation it is worth looking at the physical meaning of equation (4.34). The term ρU 2/A is always positive (because all the three terms can be only positive). Now, it can be observed that dP can be positive or negative depending on the dA and Mach number. The meaning of the sign change for the pressure differential is that the pressure can increase or decrease. It can be observed that the critical Mach number is one. If the Mach number is larger than one than dP has opposite sign of dA. If Mach number is smaller than 48 CHAPTER 4. ISENTROPIC FLOW one dP and dA have the same sign. For the subsonic branch M < 1 the term 1/(1 M 2) is positive hence − dA > 0 = dP > 0 ⇒ dA < 0 = dP < 0 ⇒ From these observations the trends are similar to those in incompressible fluid. An increase in area results in an increase of the static pressure (converting the dynamic pressure to a static pressure). Conversely, if the area decreases (as a function of x) the pressure decreases. Note that the pressure decrease is larger in compressible flow compared to incompressible flow. For the supersonic branch M > 1, the phenomenon is different. For M > 1 the term 1/1 M 2 is negative and change the character of the equation. − dA > 0 dP < 0 ⇒ dA < 0 dP > 0 ⇒ This behavior is opposite to incompressible flow behavior. For the special case of M = 1 (sonic flow) the value of the term 1 M 2 = 0 − thus mathematically dP or dA = 0. Since physically dP can increase only → ∞ in a finite amount it must that dA = 0.It must also be noted that when M = 1 occurs only when dA = 0. However, the opposite, not necessarily means that when dA = 0 that M = 1. In that case, it is possible that dM = 0 thus the diverging side is in the subsonic branch and the flow isn’t choked. The relationship between the velocity and the pressure can be observed from equation (4.28) by solving it for dU.

dP dU = (4.35) −P U From equation (4.35) it is obvious that dU has an opposite sign to dP (since the term P U is positive). Hence the pressure increases when the velocity decreases and vice versa. From the speed of sound, one can observe that the density, ρ, increases with pressure and vice versa (see equation 4.36). 1 dρ = dP (4.36) c2 It can be noted that in the derivations of the above equations (4.35 - 4.36), the equation of state was not used. Thus, the equations are applicable for any gas (perfect or imperfect gas). The second law (isentropic relationship) dictates that ds = 0 and from thermodynamics dT dP ds = 0 = C R p T − P 4.2. ISENTROPIC CONVERGING-DIVERGING FLOW IN CROSS SECTION 49 and for perfect gas dT k 1 dP = − (4.37) T k P Thus, the temperature varies according to the same way that pressure does. The relationship between the Mach number and the temperature can be obtained by utilizing the fact that the process is assumed to be adiabatic dT0 = 0. Differentiation of equation (4.9), the relationship between the temperature and the stagnation temperature becomes k 1 dT = 0 = dT 1 + − M 2 + T (k 1)MdM (4.38) 0 2 − and simplifying equation (4.38) yields dT (k 1)MdM = − (4.39) k 1 2 T − 1 + −2 M

Relationship Between the Mach Number and Cross Section Area The equations used in the solution are energy (4.39), second law (4.37), state (4.29), mass (4.26)1. Note, equation (4.33) isn’t the solution but demonstration of certain properties on the pressure. The relationship between temperature and the cross section area can be obtained by utilizing the relationship between the pressure and temperature (4.37) and the relationship of pressure and cross section area (4.33). First stage equation (4.39) is combined with equation (4.37) and becomes (k 1) dP (k 1)MdM − = − (4.40) k P k 1 2 − 1 + −2 M Combining equation (4.40) with equation (4.33) yields

ρU 2 dA 1 A 1 M 2 MdM − = (4.41) k 1 2 k P −1 + −2 M The following identify, ρU 2 = kMP can be proved as

M 2 P P U 2 U 2 kM 2P = k ρRT = k ρRT = ρU 2 (4.42) z}|{c2 kRT Using the identity in equation (4.42)zchanges}|{ equationz}|{ (4.41) into dA M 2 1 = − dM (4.43) k 1 2 A M 1 + −2 M 1The momentum equation is not used normally in isentropic process, why? 50 CHAPTER 4. ISENTROPIC FLOW Equation (4.43) is very important because it relates the geometry M, A (area) with the relative velocity (Mach number). In equation (4.43), the factors k 1 2 A, cross M 1 + −2 M and A are positive re- section

gardless of the values of M or A. There- ¡£¢

fore, the only factor that affects relation- ¡¥¤§¦©¨

ship between the cross area and the

Mach number is M 2 1. For M < 1 © − the Mach number is varied opposite to M, Much nubmer the cross section area. In the case of M > 1 the Mach number increases with the cross section area and vice versa. x The special case is when M = 1 which Fig. 4.5: The relationship between the cross requires that dA = 0. This condition section and the Mach number on the 2 imposes that internal ﬂow has to pass subsonic branch a converting–diverging device to obtain supersonic velocity. indexsonic trasition This minimum area is referred to as “throat.” Again, the opposite conclusion that when dA = 0 implies that M = 1 is not correct because possibility of dM = 0. In subsonic ﬂow branch, from the mathematical point of view: on one hand, a decrease of the cross section increases the velocity and the Mach number, on the other hand, an increase of the cross section decreases the velocity and Mach number (see Figure (4.5)).

4.2.2 Isentropic Flow Examples

Example 4.1: Air is allowed to flow from a reservoir with temperature of 21◦C and with pressure of 5[MPa] through a tube. It was measured that air mass flow rate is 1[kg/sec]. At some point on the tube static pressure was measured to be 3[MPa]. Assume that process is isentropic and neglect the velocity at the reservoir, calculate the Mach number, velocity, and the cross section area at that point where the static pressure was measured. Assume that the ratio of specific heat is k = Cp/Cv = 1.4.

SOLUTION The stagnation conditions at the reservoir will be maintained throughout the tube because the process is isentropic. Hence the stagnation temperature can be written T0 = constant and P0 = constant and both of them are known (the condition at the reservoir). For the point where the static pressure is known, the Mach number can be calculated by utilizing the pressure ratio. With the known Mach number,

2This condition does not impose any restrictions for external ﬂow. In external ﬂow, an object can be moved in arbitrary speed. 4.2. ISENTROPIC CONVERGING-DIVERGING FLOW IN CROSS SECTION 51 the temperature, and velocity can be calculated. Finally, the cross section can be calculated with all these information. In the point where the static pressure known P 3[MP a] P¯ = = = 0.6 P0 5[MP a] From Table (4.2) or from Figure (4.3) or utilizing the enclosed program from Potto- GDC, or simply using the equations shows that

M T ρ A P A P F ? × T0 ρ0 A P0 A P0 F ∗× ∗ 0.88639 0.86420 0.69428 1.0115 0.60000 0.60693 0.53105

With these values the static temperature and the density can be calculated.

T = 0.86420338 (273 + 21) = 254.076K × ρ0 ρ P 5 106[P a] ρ = 0 = 0.69428839 × ρ0 zR}|T0{ × 287.0 J 294[K] kgK × kg h i = 41.1416 m3 The velocity at that point is

c U = M √kRT = 0.88638317 √1.4 287 294 = 304[m/sec] × × × z }| { The tube area can be obtained from the mass conservation as

m˙ 5 3 A = = 8.26 10− [m ] (4.44) ρU × For a circular tube the diameter is about 1[cm].

Example 4.2: The Mach number at point A on tube is measured to be M = 23and the static pressure is 2[Bar]4. Downstream at point B the pressure was measured to be 1.5[Bar]. Calculate the Mach number at point B under the isentropic ﬂow assumption. Also, estimate the temperature at point B. Assume that the speciﬁc heat ratio k = 1.4 and assume a perfect gas model.

4This pressure is about two atmospheres with temperature of 250[K] 4Well, this question is for academic purposes, there is no known way for the author to directly measure the Mach number. The best approximation is by using inserted cone for supersonic ﬂow and measure the oblique shock. Here it is subsonic and this technique is not suitable. 52 CHAPTER 4. ISENTROPIC FLOW

SOLUTION With the known Mach number at point A all the ratios of the static properties to total (stagnation) properties can be calculated. Therefore, the stagnation pressure at point A is known and stagnation temperature can be calculated. At M = 2 (supersonic ﬂow) the ratios are

M T ρ A P A P F ? × T0 ρ0 A P0 A P0 F ∗× ∗ 2.0000 0.55556 0.23005 1.6875 0.12780 0.21567 0.59309

With this information the pressure at point B can be expressed as

from the table 4.2 @ M = 2 P P P 2.0 A = B A = 0.12780453 = 0.17040604 P0 z}|{P0 ×PB × 1.5

The corresponding Mach number for this pressure ratio is 1.8137788 and TB = 0.60315132 PB = 0.17040879. The stagnation temperature can be “bypassed” to P0 calculate the temperature at point B

M=2 M=1.81..

T0 TB 1 TB = TA = 250[K] 0.60315132 271.42[K] × z}|{TA × z}|{T0 × 0.55555556 × '

Example 4.3: Gas flows through a converging–diverging duct. At point “A” the cross section area is 50 [cm2] and the Mach number was measured to be 0.4. At point B in the duct the cross section area is 40 [cm2]. Find the Mach number at point B. Assume that the flow is isentropic and the gas specific heat ratio is 1.4.

SOLUTION To obtain the Mach number at point B by ﬁnding the ratio of the area to the critical area. This relationship can be obtained by

from the Table 4.2 A A A 40 B = B A = 1.59014 = 1.272112 A AA × A∗ 50 × ∗ z }| { AB With the value of A from the Table (4.2) or from Potto-GDC two solutions can be obtained. The two∗ possible solutions: the ﬁrst supersonic M = 1.6265306 and second subsonic M = 0.53884934. Both solution are possible and acceptable. The supersonic branch solution is possible only if there where a transition at throat where M=1. 4.2. ISENTROPIC CONVERGING-DIVERGING FLOW IN CROSS SECTION 53

M T ρ A P A P ? × T0 ρ0 A P0 A P0 ∗× 1.6266 0.65396 0.34585 1.2721 0.22617 0.28772 0.53887 0.94511 0.86838 1.2721 0.82071 1.0440

4.2.3 Mass Flow Rate (Number) One of the important engineering parameters is the mass flow rate which for ideal gas is P m˙ = ρUA = UA (4.45) RT This parameter is studied here, to examine the maximum flow rate and to see what is the effect of the compressibility on the flow rate. The area ratio as a function of the Mach number needed to be established, specifically and explicitly the relationship for the chocked flow. The area ratio is defined as the ratio of the cross section at any point to the throat area (the narrow area). It is convenient to rearrange the equation (4.45) to be expressed in terms of the stagnation properties as

f(M,k)

m˙ P P U k T 1 P k P T = 0 0 = 0 M 0 (4.46) A P0 √kRT rRr T √T0 √T0 rR zP0 r}| T{ Expressing the temperature in terms of Mach number in equation (4.46) results in

k+1 2(k 1) m˙ kMP k 1 − − = 0 1 + − M 2 (4.47) A √kRT 2 0 It can be noted that equation (4.47) holds everywhere in the converging-diverging duct and this statement also true for the throat. The throat area can be denoted as by A∗. It can be noticed that at the throat when the ﬂow is chocked or in other words M = 1 and that the stagnation conditions (i.e. temperature, pressure) do not change. Hence equation (4.47) obtained the form

k+1 2(k 1) m˙ √kP k 1 − − = 0 1 + − (4.48) A √RT 2 ∗ 0 ! Since the mass ﬂow rate is constant in the duct, dividing equations (4.48) by equation (4.47) yields

k+1 k 1 2 2(k 1) A 1 1 + − M − = 2 (4.49) A M k+1 ∗ 2 ! Equation (4.49) relates the Mach number at any point to the cross section area ratio. 54 CHAPTER 4. ISENTROPIC FLOW The maximum ﬂow rate can be expressed either by taking the derivative of equation (4.48) in with respect to M and equating to zero. Carrying this calculation results at M = 1.

k+1 2(k 1) m˙ P k k + 1 − − 0 = (4.50) A √T R 2 ∗ max 0 r For speciﬁc heat ratio, k = 1.4 m˙ P 0.68473 0 (4.51) A √T ∼ √ ∗ max 0 R

The maximum ﬂow rate for air (R = 287j/kgK) becomes,

m˙ √T 0 = 0.040418 (4.52) A∗P0 Equation (4.52) is known as Fliegner’s Formula on the name of one of the ﬁrst engineers who observed experimentally the choking phenomenon. It can be noticed that Fliengner’s equation can lead to deﬁnition of the Fliengner’s Number.

c0 F n m˙ √T m˙ kRT mc˙ 1 0 = 0 = 0 (4.53) A∗P0 √kpzRA}|∗P{0 z√R}|A∗P0{ √k The deﬁnition of Fliengner’s number (Fn) is mc˙ F n 0 (4.54) ≡ √RA∗P0

Utilizing Fliengner’s number deﬁnition and substituting it into equation (4.48) results in

k+1 2(k 1) k 1 − − F n = kM 1 + − M 2 (4.55) 2 and the maximum point for F n at M = 1 is

k+1 2(k 1) k + 1 − − F n = k (4.56) 2

“Naughty Professor” Problems in Isentropic Flow To explain the material better some instructors invented problems, which have mostly academic proposes, (see for example, Shapiro (problem 4.5)). While these 4.2. ISENTROPIC CONVERGING-DIVERGING FLOW IN CROSS SECTION 55 problems have a limit applicability in reality, they have substantial academic value and therefore presented here. The situation where the mass flow rate per area given with one of the stagnation properties and one of the static properties, e.g. P0 and T or T0 and P present difficulty for the calculations. The use of the regular isentropic Table is not possible because there isn’t variable represent this kind problems. For this kind of problems a new Table was constructed and present here5. The case of T0 and P This case considered to be simplest case and will first presented here. Using energy equation (4.9) and substituting for Mach number M = m/˙ Aρc results in

2 T0 k 1 m˙ = 1 + − (4.57) T 2 Aρc Rearranging equation (4.57) result in

1/kR p R 2 2 T k 1 m˙ T0ρ = T ρ ρ + − (4.58) z }|c2 { 2 A z}|{ And further Rearranging equation (4.58) transformed it into

P ρ k 1 m˙ 2 ρ2 = + − (4.59) T R 2kRT A 0 0 Equation (4.59) is quadratic equation for density, ρ when all other variables are known. It is convenient to change it into

P ρ k 1 m˙ 2 ρ2 − = 0 (4.60) − T R − 2kRT A 0 0 The only physical solution is when the density is positive and thus the only solution is

2 2 1  P P k 1 m˙  ρ = + + 2 − (4.61) 2 RT v RT kRT A  0 u 0 0   u   u , (M 0) 0   u → → →   t | {z } For almost incompressible ﬂow the density is reduced and the familiar form of perfect gas model is seen since stagnation temperature is approaching the static temperature for very small Mach number (ρ = P ). In other words, the terms RT0 for the group over the under–brace approaches zero when the ﬂow rate (Mach number) is very small.

5Since version 0.44 of this book. 56 CHAPTER 4. ISENTROPIC FLOW It is convenient to denote a new dimensionless density as ρ ρRT 1 ρˆ = = 0 = (4.62) p P T¯ RT0 With this new deﬁnition equation (4.61) is transformed into

1 (k 1)RT m˙ 2 ρˆ = 1 + 1 + 2 − 0 (4.63) 2  s kP 2 A    The dimensionless density now is related to a dimensionless group that is a function of Fn number and Mach number only! Thus, this dimensionless group is function of Mach number only.

2 A∗P0 F n AP =f(M)

2 2 2 2 2 RT m˙ 1 c m˙ A∗ P 0 = 0 0 (4.64) P 2 A k zP 2 }|A { z A }| P { 0 ∗ Thus,

2 2 2 RT m˙ F n A∗P 0 = 0 (4.65) P 2 A k AP Hence, the dimensionless density is

1 (k 1)F n2 A P 2 ρˆ = 1 + 1 + 2 − ∗ 0 (4.66) 2  k2 AP  s   Again notice that the right hand side of equation (4.66) is only function of Mach number (well, also the specific heat, k). And the values of AP were tabulated A∗P0 in Table (4.2) and Fn is tabulated in the next Table (4.1). Thus, the problems is reduced to finding tabulated values. The case of P0 and T A similar problem can be described for the case of stagnation pressure, P0, and static temperature, T . First, it is shown that the dimensionless group is a function of Mach number only (well, again the specific heat ratio, k also).

2 2 2 2 RT m˙ F n A∗P0 T P0 2 = (4.67) P A k AP T0 P 0 It can be noticed that F n2 T P 2 = 0 (4.68) k T P 0 4.2. ISENTROPIC CONVERGING-DIVERGING FLOW IN CROSS SECTION 57 Thus equation (4.67) became

2 2 RT m˙ A∗P = 0 (4.69) P 2 A AP 0 The right hand side is tabulated in the “regular” isentropic Table such (4.2). This example shows how a dimensional analysis is used to solve a problems without actually solving any equations. The actual solution of the equation is left as exercise (this example under construction). What is the legitimacy of this method? The explanation simply based the previous experience in which for a given ratio of area or pressure ratio (etcetera) determines the Mach number. Based on the same arguments, if it was shown that a group of parameters depends only Mach number than the Mach is determined by this group. The method of solution for given The case of ρ0 and T or P The last case sometimes referred to as the “naughty professor’s question” case dealt here is when the stagnation density given with the static temperature/pressure. First, the dimensionless approach is used later analytical method is discussed (under construction).

2 c0 2 2 2 2 2 1 m˙ kRT0 m˙ c0 m˙ F n P0 = P = 2 P = (4.70) Rρ0P A kRP0P0 A kRP0 A k P z }| { P0 P0

The last case dealt here is of the stagnation density with static pressure and the following is dimensionless group

2 c0 2 2 2 2 2 1 m˙ kRT0 T0 m˙ c0 T0 m˙ F n T0 2 = 2 = 2 = (4.71) Rρ0 T A kRP T A kRP T A k T z }| {0 0

It was hidden in the derivations/explanations of the above analysis didn’t explicitly state under what conditions these analysis is correct. Unfortunately, not all the analysis valid for the same conditions and is as the regular “isentropic” Table, (4.2). The heat/temperature part is valid for enough adiabatic condition while the pressure condition requires also isentropic process. All the above conditions/situations require to have the perfect gas model as the equation of state. For example the first “naughty professor” question is sufficient that process is adiabatic only (T0, P , mass flow rate per area.). 58 CHAPTER 4. ISENTROPIC FLOW Table 4.1: Fliegner’s number and other paramters as function of Mach number

2 2 2 2 P0A∗ RT0 m˙ 1 m˙ 1 m˙ M Fn ρˆ 2 2 AP P A Rρ0P A Rρ0 T A 0.00E+001.400E 061.000 0.0 0.0 0.0 0.0 − 0.050001 0.070106 1.000 0.00747 2.62E 05 0.00352 0.00351 − 0.10000 0.14084 1.000 0.029920 0.000424 0.014268 0.014197 0.20000 0.28677 1.001 0.12039 0.00707 0.060404 0.059212 0.21000 0.30185 1.001 0.13284 0.00865 0.067111 0.065654 0.22000 0.31703 1.001 0.14592 0.010476 0.074254 0.072487 0.23000 0.33233 1.002 0.15963 0.012593 0.081847 0.079722 0.24000 0.34775 1.002 0.17397 0.015027 0.089910 0.087372 0.25000 0.36329 1.003 0.18896 0.017813 0.098460 0.095449 0.26000 0.37896 1.003 0.20458 0.020986 0.10752 0.10397 0.27000 0.39478 1.003 0.22085 0.024585 0.11710 0.11294 0.28000 0.41073 1.004 0.23777 0.028651 0.12724 0.12239 0.29000 0.42683 1.005 0.25535 0.033229 0.13796 0.13232 0.30000 0.44309 1.005 0.27358 0.038365 0.14927 0.14276 0.31000 0.45951 1.006 0.29247 0.044110 0.16121 0.15372 0.32000 0.47609 1.007 0.31203 0.050518 0.17381 0.16522 0.33000 0.49285 1.008 0.33226 0.057647 0.18709 0.17728 0.34000 0.50978 1.009 0.35316 0.065557 0.20109 0.18992 0.35000 0.52690 1.011 0.37474 0.074314 0.21584 0.20316 0.36000 0.54422 1.012 0.39701 0.083989 0.23137 0.21703 0.37000 0.56172 1.013 0.41997 0.094654 0.24773 0.23155 0.38000 0.57944 1.015 0.44363 0.10639 0.26495 0.24674 0.39000 0.59736 1.017 0.46798 0.11928 0.28307 0.26264 0.40000 0.61550 1.019 0.49305 0.13342 0.30214 0.27926 0.41000 0.63386 1.021 0.51882 0.14889 0.32220 0.29663 0.42000 0.65246 1.023 0.54531 0.16581 0.34330 0.31480 0.43000 0.67129 1.026 0.57253 0.18428 0.36550 0.33378 0.44000 0.69036 1.028 0.60047 0.20442 0.38884 0.35361 0.45000 0.70969 1.031 0.62915 0.22634 0.41338 0.37432 0.46000 0.72927 1.035 0.65857 0.25018 0.43919 0.39596 0.47000 0.74912 1.038 0.68875 0.27608 0.46633 0.41855 0.48000 0.76924 1.042 0.71967 0.30418 0.49485 0.44215 0.49000 0.78965 1.046 0.75136 0.33465 0.52485 0.46677 0.50000 0.81034 1.050 0.78382 0.36764 0.55637 0.49249 0.51000 0.83132 1.055 0.81706 0.40333 0.58952 0.51932 0.52000 0.85261 1.060 0.85107 0.44192 0.62436 0.54733 0.53000 0.87421 1.065 0.88588 0.48360 0.66098 0.57656 0.54000 0.89613 1.071 0.92149 0.52858 0.69948 0.60706 0.55000 0.91838 1.077 0.95791 0.57709 0.73995 0.63889 4.2. ISENTROPIC CONVERGING-DIVERGING FLOW IN CROSS SECTION 59 Table 4.1: Fliegner’s number and other paramters as function of Mach number (continue)

2 2 2 2 P0A∗ RT0 m˙ 1 m˙ 1 m˙ M Fn ρˆ 2 2 AP P A Rρ0P A Rρ0 T A 0.56000 0.94096 1.083 0.99514 0.62936 0.78250 0.67210 0.57000 0.96389 1.090 1.033 0.68565 0.82722 0.70675 0.58000 0.98717 1.097 1.072 0.74624 0.87424 0.74290 0.59000 1.011 1.105 1.112 0.81139 0.92366 0.78062 0.60000 1.035 1.113 1.152 0.88142 0.97562 0.81996 0.61000 1.059 1.122 1.194 0.95665 1.030 0.86101 0.62000 1.084 1.131 1.236 1.037 1.088 0.90382 0.63000 1.109 1.141 1.279 1.124 1.148 0.94848 0.64000 1.135 1.151 1.323 1.217 1.212 0.99507 0.65000 1.161 1.162 1.368 1.317 1.278 1.044 0.66000 1.187 1.173 1.414 1.423 1.349 1.094 0.67000 1.214 1.185 1.461 1.538 1.422 1.147 0.68000 1.241 1.198 1.508 1.660 1.500 1.202 0.69000 1.269 1.211 1.557 1.791 1.582 1.260 0.70000 1.297 1.225 1.607 1.931 1.667 1.320 0.71000 1.326 1.240 1.657 2.081 1.758 1.382 0.72000 1.355 1.255 1.708 2.241 1.853 1.448 0.73000 1.385 1.271 1.761 2.412 1.953 1.516 0.74000 1.415 1.288 1.814 2.595 2.058 1.587 0.75000 1.446 1.305 1.869 2.790 2.168 1.661 0.76000 1.477 1.324 1.924 2.998 2.284 1.738 0.77000 1.509 1.343 1.980 3.220 2.407 1.819 0.78000 1.541 1.362 2.038 3.457 2.536 1.903 0.79000 1.574 1.383 2.096 3.709 2.671 1.991 0.80000 1.607 1.405 2.156 3.979 2.813 2.082 0.81000 1.642 1.427 2.216 4.266 2.963 2.177 0.82000 1.676 1.450 2.278 4.571 3.121 2.277 0.83000 1.712 1.474 2.340 4.897 3.287 2.381 0.84000 1.747 1.500 2.404 5.244 3.462 2.489 0.85000 1.784 1.526 2.469 5.613 3.646 2.602 0.86000 1.821 1.553 2.535 6.006 3.840 2.720 0.87000 1.859 1.581 2.602 6.424 4.043 2.842 0.88000 1.898 1.610 2.670 6.869 4.258 2.971 0.89000 1.937 1.640 2.740 7.342 4.484 3.104 0.90000 1.977 1.671 2.810 7.846 4.721 3.244 0.91000 2.018 1.703 2.882 8.381 4.972 3.389 0.92000 2.059 1.736 2.955 8.949 5.235 3.541 0.93000 2.101 1.771 3.029 9.554 5.513 3.699 0.94000 2.144 1.806 3.105 10.20 5.805 3.865 60 CHAPTER 4. ISENTROPIC FLOW Table 4.1: Fliegner’s number and other paramters as function of Mach number (continue)

2 2 2 2 P0A∗ RT0 m˙ 1 m˙ 1 m˙ M Fn ρˆ 2 2 AP P A Rρ0P A Rρ0 T A 0.95000 2.188 1.843 3.181 10.88 6.112 4.037 0.96000 2.233 1.881 3.259 11.60 6.436 4.217 0.97000 2.278 1.920 3.338 12.37 6.777 4.404 0.98000 2.324 1.961 3.419 13.19 7.136 4.600 0.99000 2.371 2.003 3.500 14.06 7.515 4.804 1.000 2.419 2.046 3.583 14.98 7.913 5.016

Example 4.4: A gas ﬂows in the tube with mass ﬂow rate of 1 [kg/sec] and tube cross section is 0.001[m2]. The temperature at Chamber supplying the pressure to tube is 27◦C. At some point the static pressure was measured to be 1.5[Bar]. Calculate for that point the Mach number, the velocity, and the stagnation pressure. Assume that the process is isentropic and k=1.3.

SOLUTION

The second academic condition is when the static temperature is given with the stagnation pressure. The third academic condition is of static temperature and the static pressure.

Flow with pressure losses The expression for the mass flow rate (4.47) is appropriate regardless the flow is isentropic or adiabatic. That expression was derived based on the theoretical total pressure and temperature (Mach number) which does not based on the considerations whether the flow is isentropic or adiabatic. In the same manner the definition of A∗ referred to the theoretical minimum area (”throat area”) if the flow continues to flow in an isentropic manner. Clearly, in a case where the flow isn’t isentropic or adiabatic the total pressure and the total temperature will change (due to friction, and heat transfer). A constant flow rate requires that m˙ A = m˙ B. Denoting subscript A for one point and subscript B for another point mass equation (4.48) can be equated as

k 1 2(k− 1) kP A∗ k 1 − − 0 1 + − M 2 = constant (4.72) RT 2 0

From equation (4.72), it is clear that the function f(P0, T0, A∗) = constant. There are two possible models that can be used to simplify the calculations. The first model for neglected heat transfer (adiabatic) flow and in which the total temperature remained constant (Fanno flow like). The second model which there is significant heat transfer but insignificant pressure loss (Rayleigh flow like). 4.2. ISENTROPIC CONVERGING-DIVERGING FLOW IN CROSS SECTION 61 If the mass flow rate is constant at any point on the tube (no mass loss occur) then

k+1 k 1 k 2 − m˙ = A∗ P (4.73) sRT k + 1 0 0 For adiabatic ﬂow, comparison of mass ﬂow rate at point A and point B leads to

P A∗ = P A∗ 0 |A 0 |B

P0 A∗ ; |A = |A (4.74) P A 0|B ∗|B

A∗ And utilizing the equality of A∗ = A A leads to

A A P0 A A∗ MA A | = A | (4.75) P0 B A B | A∗ MB |

For a flow with a constant stagnation pressure (frictionless flow) and non adiabatic flow reads

2 B A T0 A A∗ MB B | = A | (4.76) T0 B " A A # | A∗ MA |

Example 4.5: At point A of the tube the pressure is 3[Bar], Mach number is 2.5, and the duct section area is 0.01[m2]. Downstream at exit of tube, point B, the cross section area is 0.015[m2] and Mach number is 1.5. Assume no mass lost and adiabatic steady state ﬂow, calculated the total pressure lost.

SOLUTION Both Mach numbers are known, thus the area ratios can be calculated. The total pressure can be calculated because the Mach number and static pressure are known. With these information, and utilizing equation (4.75) the stagnation pressure at point B can be obtained.

M T ρ A P A P F ? × T0 ρ0 A P0 A P0 F ∗× ∗ 1.5000 0.68966 0.39498 1.1762 0.27240 0.32039 0.55401 2.5000 0.44444 0.13169 2.6367 0.05853 0.15432 0.62693 62 CHAPTER 4. ISENTROPIC FLOW First, the stagnation at point A is obtained from Table (4.2) as

P 3 P0 A = = = 51.25781291[Bar] | P 0.058527663

M=2.5 A

by utilizing equation (4.75)| {z pro} vides 1.1761671 0.01 P = 51.25781291 15.243[Bar] 0|B × 2.6367187 × 0.015 ≈ Hence

P P = 51.257 15.243 = 36.013[Bar] 0|A − 0|B − Note that the large total pressure loss is much larger than the static pressure loss (Pressure point B the pressure is 0.27240307 15.243 = 4.146[Bar]). × 4.3 Isentropic Tables

Table 4.2: Isentropic Table k = 1.4

M T ρ A P A P F ? × T0 ρ0 A P0 A P0 F ∗× ∗ 0.000 1.00000 1.00000 5.8E+5 1.0000 5.8E + 5 2.4E+5 0.050 0.99950 0.99875 11.59 0.99825 11.57 4.838 0.100 0.99800 0.99502 5.822 0.99303 5.781 2.443 0.200 0.99206 0.98028 2.964 0.97250 2.882 1.268 0.300 0.98232 0.95638 2.035 0.93947 1.912 0.89699 0.400 0.96899 0.92427 1.590 0.89561 1.424 0.72632 0.500 0.95238 0.88517 1.340 0.84302 1.130 0.63535 0.600 0.93284 0.84045 1.188 0.78400 0.93155 0.58377 0.700 0.91075 0.79158 1.094 0.72093 0.78896 0.55425 0.800 0.88652 0.73999 1.038 0.65602 0.68110 0.53807 0.900 0.86059 0.68704 1.009 0.59126 0.59650 0.53039 1.00 0.83333 0.63394 1.000 0.52828 0.52828 0.52828 1.100 0.80515 0.58170 1.008 0.46835 0.47207 0.52989 1.200 0.77640 0.53114 1.030 0.41238 0.42493 0.53399 1.300 0.74738 0.48290 1.066 0.36091 0.38484 0.53974 1.400 0.71839 0.43742 1.115 0.31424 0.35036 0.54655 1.500 0.68966 0.39498 1.176 0.27240 0.32039 0.55401 4.3. ISENTROPIC TABLES 63 Table 4.2: Isentropic Table k=1.4 (continue)

M T ρ A P A P F ? × T0 ρ0 A P0 A P0 F ∗× ∗ 1.600 0.66138 0.35573 1.250 0.23527 0.29414 0.56182 1.700 0.63371 0.31969 1.338 0.20259 0.27099 0.56976 1.800 0.60680 0.28682 1.439 0.17404 0.25044 0.57768 1.900 0.58072 0.25699 1.555 0.14924 0.23211 0.58549 2.000 0.55556 0.23005 1.688 0.12780 0.21567 0.59309 2.500 0.44444 0.13169 2.637 0.058528 0.15432 0.62693 3.000 0.35714 0.076226 4.235 0.027224 0.11528 0.65326 3.500 0.28986 0.045233 6.790 0.013111 0.089018 0.67320 4.000 0.23810 0.027662 10.72 0.00659 0.070595 0.68830 4.500 0.19802 0.017449 16.56 0.00346 0.057227 0.69983 5.000 0.16667 0.011340 25.00 0.00189 0.047251 0.70876 5.500 0.14184 0.00758 36.87 0.00107 0.039628 0.71578 6.000 0.12195 0.00519 53.18 0.000633 0.033682 0.72136 6.500 0.10582 0.00364 75.13 0.000385 0.028962 0.72586 7.000 0.092593 0.00261 1.0E+2 0.000242 0.025156 0.72953 7.500 0.081633 0.00190 1.4E+2 0.000155 0.022046 0.73257 8.000 0.072464 0.00141 1.9E+2 0.000102 0.019473 0.73510 8.500 0.064725 0.00107 2.5E+2 6.90E 5 0.017321 0.73723 − 9.000 0.058140 0.000815 3.3E+2 4.74E 5 0.015504 0.73903 − 9.500 0.052493 0.000631 4.2E+2 3.31E 5 0.013957 0.74058 − 10.00 0.047619 0.000495 5.4E+2 2.36E 5 0.012628 0.74192 − 4.3.1 Isentropic Isothermal Flow Nozzle 4.3.2 General Relationship In this section, the other extreme case model where the heat transfer to the gas is perfect, (e.g. Eckert number is very small) is presented. Again in reality the heat transfer is somewhere in between the two extremes. So, knowing the two limits provides a tool to examine where the reality should be expected. The perfect gas model is again assumed (later more complex models can be assumed and constructed in a future versions). In isothermal process the perfect gas model reads P = ρRT ; dP = dρRT (4.77) Substituting equation (4.77) into the momentum equation6 yields RT dP UdU + = 0 (4.78) P 6The one dimensional momentum equation for steady state is UdU/dx = dP/dx+0(other effects) − which are neglected here. 64 CHAPTER 4. ISENTROPIC FLOW Integration of equation (4.78) yields the Bernoulli’s equation for ideal gas in isothermal process which reads

U 2 U 2 P ; 2 − 1 + RT ln 2 = 0 (4.79) 2 P1 Thus, the velocity at point 2 becomes

P U = 2RT ln 2 U 2 (4.80) 2 P 1 r 1 − The velocity at point 2 for stagnation point, U 0 reads 1 ≈ P U = 2RT ln 2 (4.81) 2 P r 1 Or in explicit terms of the stagnation properties the velocity is

P U = 2RT ln (4.82) P r 0 Transform from equation (4.79) to a dimensionless form becomes constant constant 2 2 kRT (M2 M1 ) P2 ; − = R Tln (4.83) 2 P1 Simplifying equation (4.83) yields

k(M 2 M 2) P ; 2 − 1 = ln 2 (4.84) 2 P1 Or in terms of the pressure ratio equation (4.84) reads

k 2 2 k(M 2 M 2) M1 P2 1 − 2 e = e 2 = M 2 (4.85) P1 e 2 !

As oppose to the adiabatic case (T0 = constant) in the isothermal ﬂow the stagnation temperature ratio can be expressed 1 k 1 2 k 1 2 T T ¡ 1 + − M1 1 + − M1 01 = ¡1 2 = 2 (4.86) T T k 1 2 k 1 2 02 ¡ 2 1 + −2 M2 1 + −2 M2 Utilizing conservation of the mass AρM = constant to yield A M P 1 = 2 2 (4.87) A2 M1P1 4.3. ISENTROPIC TABLES 65 Combing equation (4.87) and equation (4.85) yields

k 2 2 M2 A2 M1 e = M 2 (4.88) A1 M2 e 1 ! The change in the stagnation pressure can be expressed as

k k k 1 2 2 k 1 2 − M1 P02 P2 1 + −2 M2 e = = 2 (4.89) P P k 1 2 M1 01 1 1 + −2 M1 ! "e # The critical point, at this stage, is unknown (at what Mach number the nozzle is choked is unknown) so there are two possibilities: the choking point or M = 1 to normalize the equation. Here the critical point defined as the point where M = 1 so results can be compared to the adiabatic case and denoted by star. Again it has to emphasis that this critical point is not really related to physical critical point but it is arbitrary definition. The true critical point is when flow is choked and the relationship between two will be presented. The critical pressure ratio can be obtained from (4.85) to read

P ρ (1 M2)k = = e − 2 (4.90) P ∗ ρ∗ Equation (4.88) is reduced to obtained the critical area ratio writes

A 1 (1 M2)k = e − 2 (4.91) A∗ M Similarly the stagnation temperature reads

k k 1 2 k 1 T 2 1 + − M1 − 0 = 2 (4.92) T0∗ k + 1 Finally, the critical stagnation pressure reads

k k 1 2 k 1 (1 M)k − P0 − 2 1 + −2 M1 = e 2 (4.93) P0∗ k + 1 Of course in isothermal process T = T ∗. All these equations are plotted in Figure (4.6). From the Figure 4.3 it can be observed that minimum of the curve A/A∗ isn’t on M = 1. The minimum of the curve is when area is minimum and at the point where the ﬂow is choked. It should be noted that the stagnation temperature is not constant as in the adiabatic case and the critical point is the only one constant. The mathematical procedure to ﬁnd the minimum is simply taking the derivative and equating to zero as following

k(M2 1) k(M2 1) A 2 − − d kM e 2 e 2 A∗ = 2− = 0 (4.94) dM M 66 CHAPTER 4. ISENTROPIC FLOW Isothermal Nozzle k = 1 4 4 * P / P * 3.5 A / A * P0 / P0 3 * T0 / T0 * 2.5 T / T

1.5

0.5

0 0 0.5 1 1.5 2 2.5 3 3.5 4 M

Tue Apr 5 10:20:36 2005

Fig. 4.6: Various ratios as a function of Mach number for isothermal Nozzle

Equation (4.94) simpliﬁed to

1 kM 2 1 = 0 ; M = (4.95) − √k

It can be noticed that a similar results are obtained for adiabatic ﬂow. The velocity at the throat of isothermal model is smaller by a factor of √k. Thus, dividing the critical adiabatic velocity by √k results in

√ Uthroatmax = RT (4.96)

On the other hand, the pressure loss in adiabatic ﬂow is milder as can be seen in Figure (4.7(a)). It should be emphisized that the stagnation pressure decrees. It is convenient to ﬁnd expression for the ratio of the initial stagnation pressure (the stagnation pressure before entering the nozzle) to the pressure at the throat. Utilizing equation 4.3. ISENTROPIC TABLES 67

Isothermal Nozzle Comparison between the two models k = 1 4 k = 1 4 4 5

4.5 3.5 M isoT M isentropic 4 U /U 3 isntropic isoT 3.5 2.5 3

2 2.5

* 2 1.5 A / A iso * A / A adiabatic * 1.5 1 P / P iso * P / P adiabatic 1 0.5 0.5

0 0 0 0.5 1 1.5 2 2.5 3 3.5 4 0 0.5 1 1.5 2 M Distance (normalized distance two scales)

Tue Apr 5 10:39:06 2005 Thu Apr 7 14:53:49 2005

(a) Comparison between the (b) The comparison of the adiabatic isothermal nozzle and adiabatic model and isothermal model nozzle in various variables

Fig. 4.7: The comparison of nozzle ﬂow

(4.90) the following relationship can be obtained

P P P throat = ∗ throat = P0initial P0initial P ∗

2 1 1 1 k − √k 2 = (1 02)k e„ “ ” « e −2

1 e− 2 = 0.60653 (4.97)

Notice that the critical pressure is independent of the speciﬁc heat ratio, k, as opposed to the adiabatic case. It also has to be emphasized that the stagnation values of the isothermal model are not constant. Again, the heat transfer is expressed as

Q = C (T T ) (4.98) p 02 − 02 68 CHAPTER 4. ISENTROPIC FLOW For comparison between the adiabatic model and the Comparison between the two models isothermal a simple profile of k = 1 4 nozzle area as a function of 1 the distance is assumed. This P / P profile isn’t an ideal profile 0 isentropic 0.8 T / T but rather a simple sample 0 isentropic P / P0 isothermal T/T just to examine the difference 0.6 0 isothermal between the two models so in an actual situation it can be 0.4 bounded. To make sense and eliminate unnecessary details 0.2 the distance from the entrance to the throat is normalized (to 0 0 0.5 1 1.5 2 one (1)). In the same fashion Distance (normalized distance two scales) the distance from the throat to the exit is normalized (to one Fri Apr 8 15:11:44 2005 (1)) (it doesn’t mean that these distances are the same). In this Fig. 4.8: Comparison of the pressure and temperature drop as a function of the normalized length comparison the entrance area (two scales) ratio and the exit area ratio are the same and equal to 20. The Mach number was computed for the two models and plotted in Figure (4.7(b)). In this comparison it has to be remembered that critical area for the two models are different by about 3% (for k = 1.4). As can be observed from Figure (4.7(b)). The Mach number for the isentropic is larger for the supersonic branch but the velocity is lower. The ratio of the velocities can be expressed as

U M √kRT s = s s (4.99) UT MT √kRTs

It can be noticed that temperature in the isothermal model is constant while temperature in the adiabatic model can be expressed as a function of the stagnation temperature. The initial stagnation temperatures are almost the same and can be canceled out to obtain U M s s (4.100) UT ∼ k 1 2 MT 1 + −2 Ms q By utilizing equation (4.100) the velocity ratio was obtained and is plotted in Figure (4.7(b)). Thus, using the isentropic model results in under prediction of the actual results for the velocity in the supersonic branch. While, the isentropic for the subsonic branch will be over prediction. The prediction of the Mach number are similarly shown in Figure (4.7(b)). 4.3. ISENTROPIC TABLES 69 Two other ratios need to be examined: temperature and pressure. The initial stagnation temperature is denoted as T0int. The temperature ratio of T/T0int can be obtained via the isentropic model as

T 1 = (4.101) T k 1 2 0int 1 + −2 M

While the temperature ratio of the isothermal model is constant and equal to one (1). The pressure ratio for the isentropic model is

P 1 = k 1 (4.102) P0int k 1 2 −k 1 + −2 M and for the isothermal process the stagnation pressure varies and has to be taken into account as the following:

isentropic

P P ∗ P P z = 0 0z z (4.103) P0int P0int P0∗ z}|{P0z where z is an arbitrary point on the nozzle. Using equations (4.89) and the isentropic relationship, the sought ratio is provided. Figure (4.8) shows that the range between the predicted temperatures of the two models is very large, while the range between the predicted pressure by the two models is relatively small. The meaning of this analysis is that transferred heat affects the temperature to a larger degree but the effect on the pressure is much less signiﬁcant. To demonstrate the relativity of the approach advocated in this book consider the following example.

Example 4.6: Consider a diverging–converging nozzle made out of wood (low conductive material) with exit area equal entrance area. The throat area ratio to entrance area is 1:4 respectively. The stagnation pressure is 5[Bar] and the stagnation temperature is 27◦C. Assume that the back pressure is low enough to have supersonic ﬂow without shock and k = 1.4. Calculate the velocity at the exit using the adiabatic model. If the nozzle was made from copper (a good heat conductor) a larger heat transfer occurs, should the velocity increase or decrease? What is the maximum possible increase?

SOLUTION The ﬁrst part of the question deals with the adiabatic model i.e. the conservation of the stagnation properties. Thus, with known area ratio and known stagnation Potto–GDC provides the following table: 70 CHAPTER 4. ISENTROPIC FLOW

M T ρ A P A P ? × T0 ρ0 A P0 A P0 ∗× 0.14655 0.99572 0.98934 4.0000 0.98511 3.9405 2.9402 0.36644 0.08129 4.0000 0.02979 0.11915 With the known Mach number and temperature at the exit, the velocity can be calculated. The exit temperature is 0.36644 300 = 109.9K. The exit velocity, then, × is U = M√kRT = 2.9402√1.4 287 109.9 617.93[m/sec] × × ∼ Even for the isothermal model, the initial stagnation temperature is given as 300K. Using the area ratio in Figure (4.6) or using the Potto–GDC obtains the following table

M T ρ A P A P ? × T0 ρ0 A P0 A P0 ∗× 1.9910 1.4940 0.51183 4.0000 0.12556 0.50225 The exit Mach number is known and the initial temperature to the throat temperature ratio can be calculated as the following:

T0ini 1 1 = k 1 1 = k 1 = 0.777777778 (4.104) ∗ T0 1 + −2 k 1 + −k Thus the stagnation temperature at the exit is T 0ini = 1.4940/0.777777778 = 1.921 T0exit The exit stagnation temperature is 1.92 300 = 576.2K. The exit velocity can be × determined by utilizing the following equation

U = M√kRT = 1.9910√1.4 287 300.0 = 691.253[m/sec] exit × × As was discussed before, the velocity in the copper nozzle will be larger than the velocity in the wood nozzle. However, the maximum velocity cannot exceed the 691.253[m/sec]

4.4 The Impulse Function 4.4.1 Impulse in Isentropic Adiabatic Nozzle One of the functions that is used in calculating the forces is the Impulse function. The Impulse function is denoted here as F , but in the literature some denote this function as I. To explain the motivation for using this deﬁnition consider the calculation of the net forces that acting on section shown in Figure (4.9). To calculate the net forces acting in the x–direction the momentum equation has to be applied

F = m˙ (U U ) + P A P A (4.105) net 2 − 1 2 2 − 1 1 4.4. THE IMPULSE FUNCTION 71

The net force is denoted here as Fnet. The mass conservation also can be applied to our control volume

m˙ = ρ1A1U1 = ρ2A2U2 (4.106) Combining equation (4.105) with equation (4.106) and by utilizing the identity in equation (4.42) results in F = kP A M 2 kP A M 2 P A P A (4.107) net 2 2 2 − 1 1 1 − 2 2 − 1 1 Rearranging equation (4.107) and dividing it by P0A∗ results in

f(M2) f(M1) f(M2) f(M1)

Fnet P2A2 2 P1A1 2 = 1 + kM2 1 + kM1 (4.108) P0A∗ zP0}|A∗{ − zP0}|A∗{ z }| { z }| { Examining equation (4.108) shows that the right hand side is only a function of x-direction Mach number and specific heat ratio, k. Hence, if the right hand side is only a function of the Mach number and k than the left hand side must be function of only the same parameters, M and k. Defining a function that depends only Fig. 4.9: Schematic to explain the signifi- on the Mach number creates the con- cances of the Impulse function venience for calculating the net forces acting on any device. Thus, defining the Impulse function as 2 F = P A 1 + kM2 (4.109)

In the Impulse function when F (M = 1) is denoted as F ∗

F ∗ = P ∗A∗ (1 + k) (4.110) The ratio of the Impulse function is deﬁned as see function (4.108) 2 F P1A1 1 + kM1 1 P1A1 1 = = 1 + kM 2 (4.111) P 1 F ∗ P ∗A∗ (1 + k) ∗ zP0A∗ }| { (1 + k) P0

k 2 k 1 ( k+1 ) − |{z} This ratio is different only in a coefﬁcient from the ratio deﬁned in equation (4.108) which makes the ratio a function of k and the Mach number. Hence, the net force is k k 1 k + 1 − F2 F1 F = P A∗(1 + k) (4.112) net 0 2 F − F ∗ ∗ 72 CHAPTER 4. ISENTROPIC FLOW

To demonstrate the usefulness of the this function consider a simple situation of the ﬂow through a converging nozzle

Example 4.7: 1 Consider a flow of gas into a 2 converging nozzle with a mass m˙ = 1[kg/sec] 1[kg/sec] flow rate of and the 2 2 A1 = 0.009m 2 entrance area is 0.009[m ] and A2 = 0.003m T = 400K the exit area is 0.003[m2]. The 0 P2 = 50[Bar] stagnation temperature is 400K and the pressure at point 2 was measured as 5[Bar] Calculate the net force acting on the noz- Fig. 4.10: Schematic of a flow of a compressible sub- zle and pressure at point 1. stance (gas) thorough a converging nozzle for example (4.7) SOLUTION The solution is obtained by getting the data for the Mach number. To obtained the Mach number, the ratio of P1A1/A∗P0 is needed to be calculated. To obtain this ratio the denominator is needed to be obtained. Utilizing Fliegner’s equation (4.52), provides the following m˙ √RT 1.0 √400 287 A∗P = = × × 70061.76[N] 0 0.058 0.058 ∼ and A P 500000 0.003 2 2 = × 2.1 A∗P0 70061.76 ∼

M T ρ A P A P F ? × T0 ρ0 A P0 A P0 F ∗× ∗ 0.27353 0.98526 0.96355 2.2121 0.94934 2.1000 0.96666

A With the area ratio of A? = 2.2121 the area ratio of at point 1 can be calculated.

A1 A2 A1 0.009 ? = ? = 2.2121 = 5.2227 A A A2 × 0.003 And utilizing again Potto-GDC provides

M T ρ A P A P F ? × T0 ρ0 A P0 A P0 F ∗× ∗ 0.11164 0.99751 0.99380 5.2227 0.99132 5.1774 2.1949

The pressure at point 1 is

P0 P1 P1 = P2 = 5.0times0.94934/0.99380 4.776[Bar] P2 P0 ∼ 4.5. ISOTHERMAL TABLE 73 The net force is obtained by utilizing equation (4.112)

k k 1 P A k + 1 − F F F = P A 0 ∗ (1 + k) 2 1 net 2 2 P A 2 F − F 2 2 ∗ ∗ 1 = 500000 2.4 1.23.5 (2.1949 0.96666) 614[kN] × 2.1 × × × − ∼

4.4.2 The Impulse Function in Isothermal Nozzle Previously Impulse function was developed in the isentropic adiabatic flow. The same is done here for the isothermal nozzle flow model. As previously, the definition of the Impulse function is reused. The ratio of the impulse function for two points on the nozzle is

2 F2 P2A2 + ρ2U2 A2 = 2 (4.113) F1 P1A1 + ρ1U1 A1 Utilizing the ideal gas model for density and some rearrangement results in

2 F P A 1 + U2 2 = 2 2 RT U 2 (4.114) F1 P1A1 1 1 + RT

Since U 2/RT = kM 2 and the ratio of equation (4.87) transformed equation into (4.114)

2 F2 M1 1 + kM2 = 2 (4.115) F1 M2 1 + kM1 At the star condition (M = 1) (not the minimum point) results in

2 F2 1 1 + kM2 = (4.116) F ∗ M2 1 + k

4.5 Isothermal Table

Table 4.3: Isothermal Table

M T0 P0 A P A P F ? ? ? ? × T0 P0 A P A P0 F ∗× ∗ 0.00 0.52828 1.064 5.0E + 5 2.014 1.0E+6 4.2E+5 0.05 0.52921 1.064 9.949 2.010 20.00 8.362 0.1 0.53199 1.064 5.001 2.000 10.00 4.225 74 CHAPTER 4. ISENTROPIC FLOW Table 4.3: Isothermal Table (continue)

M T0 P0 A P A P F ? ? ? ? × T0 P0 A P A P0 F ∗× ∗ 0.2 0.54322 1.064 2.553 1.958 5.000 2.200 0.3 0.56232 1.063 1.763 1.891 3.333 1.564 0.4 0.58985 1.062 1.389 1.800 2.500 1.275 0.5 0.62665 1.059 1.183 1.690 2.000 1.125 0.6 0.67383 1.055 1.065 1.565 1.667 1.044 0.7 0.73278 1.047 0.99967 1.429 1.429 1.004 0.8 0.80528 1.036 0.97156 1.287 1.250 0.98750 0.9 0.89348 1.021 0.97274 1.142 1.111 0.98796 1.00 1.000 1.000 1.000 1.000 1.000 1.000 1.10 1.128 0.97376 1.053 0.86329 0.90909 1.020 1.20 1.281 0.94147 1.134 0.73492 0.83333 1.047 1.30 1.464 0.90302 1.247 0.61693 0.76923 1.079 1.40 1.681 0.85853 1.399 0.51069 0.71429 1.114 1.50 1.939 0.80844 1.599 0.41686 0.66667 1.153 1.60 2.245 0.75344 1.863 0.33554 0.62500 1.194 1.70 2.608 0.69449 2.209 0.26634 0.58824 1.237 1.80 3.035 0.63276 2.665 0.20846 0.55556 1.281 1.90 3.540 0.56954 3.271 0.16090 0.52632 1.328 2.00 4.134 0.50618 4.083 0.12246 0.50000 1.375 2.50 9.026 0.22881 15.78 0.025349 0.40000 1.625 3.000 19.41 0.071758 90.14 0.00370 0.33333 1.889 3.500 40.29 0.015317 7.5E + 2 0.000380 0.28571 2.161 4.000 80.21 0.00221 9.1E + 3 2.75E 5 0.25000 2.438 − 4.500 1.5E + 2 0.000215 1.6E + 5 1.41E 6 0.22222 2.718 − 5.000 2.8E + 2 1.41E 5 4.0E + 6 0.0 0.20000 3.000 − 5.500 4.9E + 2 0.0 1.4E + 8 0.0 0.18182 3.284 6.000 8.3E + 2 0.0 7.3E + 9 0.0 0.16667 3.569 6.500 1.4E + 3 0.0 5.3E+11 0.0 0.15385 3.856 7.000 2.2E + 3 0.0 5.6E+13 0.0 0.14286 4.143 7.500 3.4E + 3 0.0 8.3E+15 0.0 0.13333 4.431 8.000 5.2E + 3 0.0 1.8E+18 0.0 0.12500 4.719 8.500 7.7E + 3 0.0 5.4E+20 0.0 0.11765 5.007 9.000 1.1E + 4 0.0 2.3E+23 0.0 0.11111 5.296 9.500 1.6E + 4 0.0 1.4E+26 0.0 0.10526 5.586 10.00 2.2E + 4 0.0 1.2E+29 0.0 0.100000 5.875

4.6 The effects of Real Gases To obtained expressions for non–ideal gas it is communally done by reusing the ideal gas model and introducing a new variable which is a function of the gas prop- 4.6. THE EFFECTS OF REAL GASES 75 erties like the critical pressure and critical temperature. Thus, a real gas equation can be expressed in equation (3.19). Differentiating equation (3.19) and dividing by equation (3.19) yields dP dz dρ dT = + + (4.117) P z ρ T Again, Gibb’s equation (4.27) is reused to related the entropy change to the change in thermodynamics properties and applied on non-ideal gas. Since ds = 0 and utilizing the equation of the state dh = dP/ρ. The enthalpy is a function of the temperature and pressure thus, h = h(T, P ) and full differential is ∂h ∂h dh = dT + dP (4.118) ∂T ∂P P T The definition of pressure specific heat is C ∂h and second derivative is p ≡ ∂T Maxwell relation hence, ∂h ∂s = v T (4.119) ∂P − ∂T T P First, the differential of enthalpy is calculated for real gas equation of state as T ∂z dP dh = C dT (4.120) p − Z ∂T ρ P Equations (4.27) and (3.19) are combined to form ds C dT T ∂z dP = p z 1 + (4.121) R R T − Z ∂T P P The mechanical energy equation can be expressed as U 2 dP d = (4.122) 2 − ρ Z Z At the stagnation the definition requires that the velocity is zero. To carry the integration of the right hand side the relationship between the pressure and the density has to be defined. The following power relationship is assumed

1 ρ P n = (4.123) ρ P 0 0 Notice, that for perfect gas the n is substituted by k. With integration of equation (4.122) when using relationship which is deﬁned in equation (4.123) results

1 U 2 P1 dP P 1 P n = = 0 dP (4.124) 2 ρ ρ P ZP0 ZP0 0 76 CHAPTER 4. ISENTROPIC FLOW Substituting relation for stagnation density (3.19) results

1 2 P n U z0RT0 P0 = dP (4.125) 2 P P ZP0 0 For n > 1 the integration results in

n 1 2n P ( −n ) U = z RT 1 (4.126) v 0 0 u n 1 " − P0 # u − t For n = 1 the integration becomes

P U = 2z RT ln 0 (4.127) 0 0 P s It must be noted that n is a function of the critical temperature and critical pressure. The mass ﬂow rate is regardless to equation of state as following

m˙ = ρ∗A∗U ∗ (4.128)

Where ρ∗ is the density at the throat (assuming the chocking condition) and A∗ is the cross area of the throat. Thus, the mass ﬂow rate in our properties

U ∗ ρ∗

1 n 1 n ( −n ) P0 P 2n P m˙ = A∗ z RT 1 (4.129) vz 0 0 }| { zz0RT0}| P0 { u n 1 " − P0 # u − t For the case of n = 1

ρ∗ U ∗∗ 1 n P0 P P0 m˙ = A∗ 2z RT ln (4.130) zz RT }| P { z 0 }|0 P { 0 0 0 s The Mach number can be obtained by utilizing equation (3.34) to deﬁned the Mach number as U M = (4.131) √znRT

Integrating equation (4.121) when ds = 0 results

T2 C dT P2 T ∂z dP p = z 1 + (4.132) R T Z ∂T P ZT1 ZP1 P 4.6. THE EFFECTS OF REAL GASES 77 To carryout the integration of equation (4.132) looks at Bernnolli’s equation which is dU 2 dP = (4.133) 2 − ρ Z Z After integration of the velocity

2 P/P0 dU ρ0 P = d (4.134) 2 − ρ P Z1 0 It was shown in Chapter (3) that (3.33) is applicable for some ranges of relative temperature and pressure (relative to critical temperature and pressure and not the stagnation conditions).

n 1 2n P −n U = z RT 1 (4.135) v 0 0 u n 1 " − P0 # u − t When n = 1 or when n 1 → P U = 2z RT ln 0 (4.136) 0 0 P s

The mass ﬂow rate for the real gas m˙ = ρ∗U ∗A∗

1 n A∗P 2n P ∗ P ∗ m˙ = 0 1 (4.137) √z RT n 1 P − P 0 0 r − 0 0 And for n = 1

A∗P0 2n P0 m˙ = 2z0RT0 ln (4.138) √z RT n 1 P 0 0 r − s

Fliegner’s number in this case is

1 n mc˙ 2n P ∗ P ∗ F n = 0 1 (4.139) A P n 1 P − P ∗ 0 r − 0 0

Fliegner’s number for n = 1 is

2 mc˙ P ∗ P ∗ F n = 0 = 2 ln (4.140) A P P − P ∗ 0 0 0 78 CHAPTER 4. ISENTROPIC FLOW The critical ratio of the pressure is

n n 1 P ∗ 2 − = (4.141) P n + 1 0 When n = 1 or more generally when n 1 this is a ratio approach → P ∗ = √e (4.142) P0

To obtain the relationship between the temperature and pressure, equation (4.132) can be integrated

R [z+T ( ∂z ) ] T P Cp ∂T P 0 = 0 (4.143) T P k 1 The power of the pressure ratio is approaching −k when z approaches 1. Note that

1 n T z P −n 0 = 0 0 (4.144) T z P The Mach number at every point at the nozzle can be expressed as

1 n 2 z T P 0 −n M = 0 0 1 − (4.145) v u n 1 z T " − P # u − t For n = 1 the Mach number is z T P M = 2 0 0 ln 0 (4.146) r z T P The pressure ratio at any point can be expressed as a function of the Mach number as

n 1 ∂z ( − )[z+T ( ) ] T n 1 n ∂T P 0 = 1 + − M 2 (4.147) T 2 for n = 1

2 ∂z T0 M [z+T ( ) ] = ∂T P (4.148) T e The critical temperature is given by

n ∂z ( 1 n )[z+T ( ∂T ) ] T ∗ 1 + n − P = (4.149) T 2 0 4.6. THE EFFECTS OF REAL GASES 79 and for n = 1

∂z T ∗ [z+T ( ) ] = e− ∂T P (4.150) T0 q

The mass ﬂow rate as a function of the Mach number is

n+1 n 1 P n n 1 − m˙ = 0 M 1 + − M 2 (4.151) c s 2 0 For the case of n = 1 the mass ﬂow rate is

n+1 2 n 1 P A∗n M n 1 − m˙ = 0 1 + − M 2 (4.152) c e s 2 0 q

Example 4.8: A design is required that at a speciﬁc point the Mach number should be M = 2.61, the pressure 2[Bar], and temperature 300K . i. Calculate the area ratio between the point and the throat. ii. Calculate the stagnation pressure and the stagnation temperature. iii. Are the stagnation pressure and temperature at the entrance different from the point? You can assume that k = 1.405.

SOLUTION

1. The solution is simpliﬁed by using Potto-GDC for M = 2.61 the results are

M T ρ A P A P ? × T0 ρ0 A P0 A P0 ∗× 2.6100 0.42027 0.11761 2.9066 0.04943 0.14366

2. The stagnation pressure is obtained from P 2.61 P = 0 P = 52.802[Bar] 0 P 0.04943 ∼ The stagnation temperature is T 300 T = 0 T = 713.82K 0 T 0.42027 ∼

3. Of course, the stagnation pressure is constant for isentropic ﬂow. 80 CHAPTER 4. ISENTROPIC FLOW CHAPTER 5

Normal Shock

In this chapter the relationships between the two sides of normal shock are presented. In this discussion, the ﬂow is assumed to be in a steady state, and the thickness of the shock is assumed to be very small. A discussion on the shock thickness will be presented in a forthcoming section1. A shock can occur in at least two different mechanisms.

The ﬁrst is when a large differ- flow direction

ence (above a small minimum

!#" value) between the two sides of a membrane, and when the membrane bursts (see the discussion about the shock tube). c.v. Of course, the shock travels from the high pressure to the Fig. 5.1: A shock wave inside a tube, but it can also be low pressure side. The sec- viewed as a one–dimensional shock wave. ond is when many sound waves “run into” each other and accumulate (some refer to it as “coalescing”) into a large difference, which is the shock wave. In fact, the sound wave can be viewed as an extremely weak shock. In the speed of sound analysis, it was assumed the medium is continuous, without any abrupt changes. This assumption is no longer valid in the case of a shock. Here, the relationship for a perfect gas is constructed. In Figure (5.1) a control volume for this analysis is shown, and the gas ﬂows from left to right. The conditions, to the left and to the right of the shock, are

1Currently under construction.

81 82 CHAPTER 5. NORMAL SHOCK assumed to be uniform2. The conditions to the right of the shock wave are uniform, but different from the left side. The transition in the shock is abrupt and in a very narrow width. The chemical reactions (even condensation) are neglected, and the shock occurs at a very narrow section. Clearly, the isentropic transition assumption is not appropriate in this case because the shock wave is a discontinued area. Therefore, the increase of the entropy is fundamental to the phenomenon and the understanding of it. It is further assumed that there is no friction or heat loss at the shock (because the heat transfer is negligible due to the fact that it occurs on a relatively small surface). It is customary in this ﬁeld to denote x as the upstream condition and y as the downstream condition. The mass ﬂow rate is constant from the two sides of the shock and therefore the mass balance is reduced to

ρxUx = ρyUy (5.1) In a shock wave, the momentum is the quantity that remains constant because there are no external forces. Thus, it can be written that P P = ρ U 2 ρ U 2 (5.2) x − y x y − y x The process is adiabatic, or nearly adiabatic , and theref ore the energy equation can be written as U 2 U 2 C T + x = C T + y (5.3) p x 2 p y 2 The equation of state for perfect gas reads P = ρRT (5.4)

If the conditions upstream are known, then there are four unknown conditions downstream. A system of four unknowns and four equations is solvable. Nevertheless, one can note that there are two solutions because of the quadratic of equation (5.3). These two possible solutions refer to the direction of the flow. Physics dictates that there is only one possible solution. One cannot deduce the direction of the flow from the pressure on both sides of the shock wave. The only tool that brings us to the direction of the flow is the second law of thermodynamics. This law dictates the direction of the flow, and as it will be shown, the gas flows from a supersonic flow to a subsonic flow. Mathematically, the second law is expressed by the entropy. For the adiabatic process, the entropy must increase. In mathematical terms, it can be written as follows: s s > 0 (5.5) y − x 2Clearly the change in the shock is so significant compared to the changes in medium before and after the shock that the changes in the mediums (flow) can be considered uniform. 83 Note that the greater–equal signs were not used. The reason is that the process is irreversible, and therefore no equality can exist. Mathematically, the parameters are P, T, U, and ρ, which are needed to be solved. For ideal gas, equation (5.5) is

T P ln y (k 1) y > 0 (5.6) Tx − − Px

It can also be noticed that entropy, s, can be expressed as a function of the other parameters. Now one can view these equations as two different subsets of equations. The first set is the energy, continuity, and state equations, and the second set is the momentum, continuity, and state equations. The solution of every set of these equations produces one additional degree of freedom, which will produce a range of possible solutions. Thus, one can have a whole range of solutions. In the first case, the energy equation is used, producing various resistance to the flow. This case is called Fanno flow, and Chapter (9) deals extensively with this topic. The mathematical explanation is given Chapter (9) in greater detail. In- stead of solving all the equations that were presented, one can solve only four (4) equations (including the second law), which will require additional parameters. If the energy, continuity, and state equations are solved for the arbitrary value of the Ty, a parabola in the T –s diagram will be obtained. On the other hand, when the momentum equation is solved instead of the energy equation, the degree of freedom is now energy, i.e., the energy amount “added” to the shock. This situation is similar to a frictionless flow with the addition of heat, and this flow is known as Rayleigh flow. This flow is dealt with in greater detail in Chapter (10).

Since the shock has $&%('

no heat transfer (a special EGFIH

K J subsonic case of Rayleigh ﬂow) and flow supersonic there isn’t essentially any ,¥-/.0.213-/.546- flow momentum transfer (a spe- T =?>A@ cial case of Fanno ﬂow), shock jump the intersection of these B?CAD two curves is what really Rayleigh Fanno line happened in the shock. In line

Figure (5.2), the intersec- )&*(+

tion is shown and two solu- 7¥8:95;38:95<68 tions are obtained. Clearly, the increase of the entropy determines the direction of s the flow. The entropy increases from point x to Fig. 5.2: The intersection of Fanno flow and Rayleigh flow point y. It is also worth not- produces two solutions for the shock wave. ing that the temperature at M = 1 on Rayleigh flow is larger than that on the Fanno line. 84 CHAPTER 5. NORMAL SHOCK 5.1 Solution of the Governing Equations 5.1.1 Informal Model Accepting the fact that the shock is adiabatic or nearly adiabatic requires that total energy is conserved, T0x = T0y. The relationship between the temperature and the stagnation temperature provides the relationship of the temperature for both sides of the shock. T y k 1 2 T T0 1 + − Mx y = y = 2 (5.7) Tx k 1 2 Tx 1 + − My T0x 2

All the other relationships are essentially derived from this equation. The only issue left to derive is the relationship between Mx and My. Note that the Mach number is a function of temperature, and thus for known Mx all the other quantities can be determined, at least, numerically. The analytical solution is discussed in the next section.

5.1.2 Formal Model Equations (5.1), (5.2), and (5.3) can be converted into a dimensionless form. The reason that dimensionless forms are heavily used in this book is because by doing so it simpliﬁes and clariﬁes the solution. It can also be noted that in many cases the dimensionless equations set is more easily solved. From the continuity equation (5.1) substituting for density, ρ, the equation of state yields

Px Py Ux = Uy (5.8) RTx RTy Squaring equation (5.8) results in P 2 P 2 x U 2 = y U 2 (5.9) 2 2 x 2 2 y R Tx R Ty Multiplying the two sides by the ratio of the specific heat, k, provides a way to obtain the speed of sound definition/equation for perfect gas, c2 = kRT to be used for the Mach number definition, as follows: 2 2 Px 2 Py 2 Ux = Uy (5.10) Tx kRTx Ty kRTy

2 2 cx cy Note that the speed of sound |on{zthe} different |sides{z } of the shock is different. Utiliz- ing the deﬁnition of Mach number results in 2 2 Px 2 Py 2 Mx = My (5.11) Tx Ty 5.1. SOLUTION OF THE GOVERNING EQUATIONS 85 Rearranging equation (5.11) results in

T P 2 M 2 y = y y (5.12) T P M x x x Energy equation (5.3) can be converted to a dimensionless form which can be expressed as

k 1 k 1 T 1 + − M 2 = T 1 + − M 2 (5.13) y 2 y x 2 x It can also be observed that equation (5.13) means that the stagnation temper- 2 ature is the same, T0y = T0x. Under the perfect gas model, ρU is identical to kP M 2 because

M 2 ρ P U 2 ρU 2 = z }| { kRT = kP M2 (5.14) z}|{RT kRT  2   c    Using the identity (5.14) transforms the| {zmomentum} equation (5.2) into

2 2 Px + kPxMx = Py + kPyMy (5.15)

Rearranging equation (5.15) yields

2 Py 1 + kMx = 2 (5.16) Px 1 + kMy

The pressure ratio in equation (5.16) can be interpreted as the loss of the static pressure. The loss of the total pressure ratio can be expressed by utilizing the relationship between the pressure and total pressure (see equation (4.11)) as

k k 1 2 k 1 P0y Py 1 + −2 My − = k (5.17) P0x k 1 2 k 1 Px 1 + −2 Mx − The relationship between Mx and My is needed to be solved from the above set of equations. This relationship can be obtained from the combination of mass, momentum, and energy equations. From equation (5.13) (energy) and equation (5.12) (mass) the temperature ratio can be eliminated.

2 k 1 2 PyMy 1 + −2 Mx = k 1 2 (5.18) PxMx 1 + M −2 y 86 CHAPTER 5. NORMAL SHOCK Combining the results of (5.18) with equation (5.16) results in

2 2 2 k 1 2 1 + kMx Mx 1 + −2 Mx 2 = k 1 2 (5.19) 1 + kM My 1 + M y −2 y

Equation (5.19) is a symmetrical equation in the sense that if My is substituted with Mx and Mx substituted with My the equation remains the same. Thus, one solution is

My = Mx (5.20)

It can be observed that equation (5.19) is biquadratic. According to the Gauss Biquadratic Reciprocity Theorem this kind of equation has a real solution in a certain range3 which will be discussed later. The solution can be obtained by rewriting equation (5.19) as a polynomial (fourth order). It is also possible to cross–multiply equation (5.19) and divide it by M 2 M 2 y − y k 1 1 + − M 2 + M 2 kM 2M 2 = 0 (5.21) 2 y y − y y Equation (5.21) becomes

2 2 2 Mx + k 1 − My = 2k 2 (5.22) k 1 Mx 1 − − The ﬁrst solution (5.20) is the trivial solution in which the two sides are identical and no shock wave occurs. Clearly, in this case, the pressure and the temperature from both sides of the nonexistent shock are the same, i.e. Tx = Ty, Px = Py. The second solution is where the shock wave occurs. The pressure ratio between the two sides can now be as a function of only a single Mach number, for example, Mx. Utilizing equation (5.16) and equation (5.22) provides the pressure ratio as only a function of the upstream Mach number as

Py 2k 2 k 1 = Mx − Px k + 1 − k + 1 Py 2k 2 = 1 + Mx 1 (5.23) Px k + 1 − The density and upstream Mach number relationship can be obtained in the same fashion to became

2 ρy Ux (k + 1)Mx = = 2 (5.24) ρx Uy 2 + (k 1)M − x 3Ireland, K. and Rosen, M. ”Cubic and Biquadratic Reciprocity.” Ch. 9 in A Classical Introduction to Modern Number Theory, 2nd ed. New York: Springer-Verlag, pp. 108-137, 1990. 5.1. SOLUTION OF THE GOVERNING EQUATIONS 87

The fact that the pressure ratio is a function of the upstream Mach number, Mx, provides additional way of obtaining an additional useful relationship. And the temperature ratio, as a function of pressure ratio, is transformed into

k+1 Py Ty Py k 1 + P = − x (5.25) +1 Py Tx Px k 1 + k 1 P ! − x In the same way, the relationship between the density ratio and pressure ratio is

1 + k+1 Py ρ k 1 Px x = − (5.26) ρy k+1 Py k 1 + P − x which is associated with the shock wave.

The Maximum Conditions Shock Wave relationship The maximum speed of sound My and P0y/P0x as a function of Mx is when the highest tempera- 1 ture is achieved. The maxi- 0.9 mum temperature that can be 0.8 My achieved is the stagnation tem- P /P 0.7 0y 0x perature 0.6 y

2k M 0.5 Umax = RT0 (5.27) k 1 0.4 r − 0.3 The stagnation speed of sound is 0.2 0.1

c0 = kRT0 (5.28) 0 1 2 3 4 5 6 7 8 9 10 M Fri Jun 18 15:47:34 2004 x Based on thisp deﬁnition a new Mach number can be deﬁned Fig. 5.3: The exit Mach number and the stagnation pressure ratio as a function of upstream Mach U M0 = (5.29) number. c0

The Star Conditions The speed of sound at the critical condition can also be a good reference velocity. The speed of sound at that velocity is

c∗ = √kRT ∗ (5.30) 88 CHAPTER 5. NORMAL SHOCK In the same manner, an additional Mach number can be deﬁned as

U M ∗ = (5.31) c∗

5.1.3 Prandtl’s Condition

It can be easily observed that the temperature from both sides of the shock wave is discontinuous. Therefore, the speed of sound is different in these adjoining mediums. It is therefore convenient to deﬁne the star Mach number that will be independent of the speciﬁc Mach number (independent of the temperature).

U c U c M ∗ = = = M (5.32) c∗ c∗ c c∗

The jump condition across the shock must satisfy the constant energy.

2 2 2 2 c U c∗ c∗ k + 1 2 + = + = c∗ (5.33) k 1 2 k 1 2 2(k 1) − − − Dividing the mass equation by the momentum equation and combining it with the perfect gas model yields

2 2 c1 c2 + U1 = + U2 (5.34) kU1 kU2

Combining equation (5.33) and (5.34) results in

1 k + 1 2 k 1 1 k + 1 2 k 1 c∗ − U + U = c∗ − U + U (5.35) kU 2 − 2 1 1 kU 2 − 2 2 2 1 2 After rearranging and diving equation (5.35) the following can be obtained:

2 U1U2 = c∗ (5.36)

or in a dimensionless form

2 M ∗1M ∗2 = c∗ (5.37) 5.2. OPERATING EQUATIONS AND ANALYSIS 89 5.2 Operating Equations and Analysis

In Figure (5.3), the Mach number after the shock, My, and the ratio of the total pressure, P0y/P0x, are plotted as a function of the entrance Mach number. The working equations were presented earlier. Note that the My has a minimum value which depends on the specific heat ratio. It can be noticed that the density ratio (velocity ratio) also has a finite value regardless of the upstream Mach number. The typical situations in which these equations can be used also include the moving shocks. The equations should be used with the Mach number (upstream or downstream) for a given pressure ratio or density ratio (velocity ratio). This kind of equations requires examining Table (5.1) for k = 1.4 or utilizing Potto-GDC for for value of the specific heat ratio. Finding the Mach number for a pressure ratio of 8.30879 and k = 1.32 is only a few mouse clicks away from the following table.

Ty ρy Py P0y Mx My Tx ρx Px P0x 2.7245 0.47642 2.1110 3.9360 8.3088 0.38109

To illustrate the use of the Shock Wave relationship P /P , ρ /ρ and T /T as a function of M above equations, an exam- y y y x y x x 120.0 ple is provided. 110.0 100.0 Py/Px Example 5.1: 90.0 Ty/Tx Air ﬂows with a Mach num- 80.0 ρy/ρx ber of Mx = 3, at a pres- 70.0 sure of 0.5 [bar] and a 60.0 temperature of 0◦C goes 50.0 through a normal shock. 40.0 Calculate the temperature, 30.0 pressure, total pressure, 20.0 and velocity downstream of 10.0 the shock. 0.0 1 2 3 4 5 6 7 8 9 10 M Fri Jun 18 15:48:25 2004 x SOLUTION Analysis: Fig. 5.4: The ratios of the static properties of the two sides First, the known informa- of the shock. tion are Mx = 3, px = 1.5[bar] and Tx = 273K. Using these data, the total pressure can be obtained (through an isentropic relationship in Table (4.2), i.e., P0x is known). Also with the temperature, Tx, the velocity can readily be calculated. The relationship that was calculated will be utilized to obtain the ratios for the downstream of the normal Px shock. = 0.0272237 = P0x = 1.5/0.0272237 = 55.1[bar] P0x ⇒ c = √kRT = √1.4 287 273 = 331.2m/sec x x × × 90 CHAPTER 5. NORMAL SHOCK

Ty ρy Py P0y Mx My Tx ρx Px P0x 3.0000 0.47519 2.6790 3.8571 10.3333 0.32834

U = M c = 3 331.2 = 993.6[m/sec] x x × x × Now the velocity downstream is determined by the inverse ratio of ρy/ρx = Ux/Uy = 3.85714.

Uy = 993.6/3.85714 = 257.6[m/sec]

P P = 0y P = 0.32834 55.1[bar] = 18.09[bar] 0y P × 0x × 0x

5.2.1 The Limitations of the Shock Wave When the upstream Mach number becomes very large, the downstream Mach number (see equation (5.22)) is limited by

: 0 2 ∼ 1 + 2 2 (k 1)Mx k 1 − My = 0 = − (5.38) 2k 1>∼ 2k k 1 M 2 − − x This result is shown in Figure (5.3). The limits of the pressure ratio can be obtained by looking at equation (5.16) and by utilizing the limit that was obtained in equation (5.38).

5.2.2 Small Perturbation Solution The small perturbation solution refers to an analytical solution where only a small change (or several small changes) occurs. In this case, it refers to a case where only a “small shock” occurs, which is up to Mx = 1.3. This approach had a major significance and usefulness at a time when personal computers were not available. Now, during the writing of this version of the book, this technique is used mostly in obtaining analytical expressions for simplified models. This technique also has an academic value and therefore will be described in the next version (0.5 series). The strength of the shock wave is defined as P P P ˆ = y − x = y 1 (5.39) P Px Px − By using equation (5.23) transforms equation (5.39) into 2k ˆ = M 2 1 (5.40) P k + 1 x − 5.3. THE MOVING SHOCKS 91 or by utilizing equation (5.24) the following is obtained:

2k ρy k 1 ρx 1 ˆ = − − (5.41) P 2 ρy k 1 ρ 1 − − x −

5.2.3 Shock Thickness The issue of shock thickness (which will be presented in a later version) is presented here for completeness. This issue has a very limited practical application for most students; however, to convince the students that indeed the assumption of very thin shock is validated by analytical and experimental studies, the issue should be presented. The shock thickness can be defined in several ways. The most common definition is by passing a tangent to the velocity at the center and finding out where the theoretical upstream and downstream conditions are meet.

5.3 The Moving Shocks In some situations, the shock wave is

not stationary. This kind of situation

[]\ ^]_

arises in many industrial applications. flow

L5M Q5R

direction g

For example, when a valve is suddenly hji

NPO

ST dfe

4 closed and a shock propagates up- `ba c

UWV XZY stream. On the other extreme, when a valve is suddenly opened or a membrane is ruptured, a shock occurs and c.v. propagates downstream (the opposite Stationary Coordinates direction of the previous case). In some industrial applications, a liquid

(metal) is pushed in two rapid stages

: {p s to a cavity through a pipe system. This x:y{zp|}y~¥

liquid (metal) is pushing gas (mostly)

air, which creates two shock stages. k#lnmpo qsrutvqpw

As a general rule, the shock can move b downstream or upstream. The last situation is the most general case, which this section will be dealing with. c.v. There are more genera cases where Moving Coordinates the moving shock is created which include a change in the physical prop- Fig. 5.5: Comparison between stationary shock and moving shock in ducts erties, but this book will not deal with

4It will be explained using dimensional analysis what is suddenly open 92 CHAPTER 5. NORMAL SHOCK them at this stage. The reluctance to deal with the most general case is due to fact it is highly specialized and complicated even beyond early graduate students level. In these changes (of opening a valve and closing a valve on the other side) create situations in which different shocks are moving in the tube. The general case is where two shocks collide into one shock and moves upstream or downstream is the general case. A speciﬁc example is common in die–casting: after the ﬁrst shock moves a second shock is created in which its velocity is dictated by the upstream and downstream velocities. In cases where the shock velocity can be approximated as a constant (in the ma- jority of cases) or as near constant, the previous analysis, equations, and the tools developed in this chapter can be employed. The problem can be reduced to the previously studied shock, i.e., to the stationary case when the coordinates are attached to the shock front. In such a case, the steady state is obtained in the moving control value. For this analysis, the coordinates move with the shock. Here, the prime ’ denote the values of the static coordinates. Note that this notation is contrary to the con- ventional notation found in the literature. The reason for the deviation is that this choice reduces the programing work (especially for object–oriented programing like C++). An observer moving with the shock will notice that the pressure in the shock is

0 0 Px = Px Py = Py (5.42) The temperature measured by the observer is

0 0 Tx = Tx Ty = Ty (5.43) Assuming that the shock is moving to the right, (refer to Figure (5.5)) the velocity measured by the observer is U = U U 0 (5.44) x s − x Where Us is the shock velocity which is moving to the right. The “downstream” velocity is U 0 = U U (5.45) y s − y The speed of sound on both sides of the shock depends only on the temperature and it is assumed to be constant. The upstream prime Mach number can be deﬁned as

0 Us Ux Us Mx = − = Mx = Msx Mx (5.46) cx cx − − It can be noted that the additional deﬁnition was introduced for the shock upstream Mach number, M = Us . The downstream prime Mach number can be expressed sx cx as

0 Us Uy Us My = − = My = Msy My (5.47) cy cy − − 5.3. THE MOVING SHOCKS 93 Similar to the previous case, an additional deﬁnition was introduced for the shock downstream Mach number, Msy. The relationship between the two new shock Mach numbers is U c U s = y s cx cx cy T M = y M (5.48) sx T sy r x The “upstream” stagnation temperature of the ﬂuid is k 1 T = T 1 + − M 2 (5.49) 0x x 2 x and the “upstream” prime stagnation pressure is

k k 1 k 1 − P = P 1 + − M 2 (5.50) 0x x 2 x The same can be said for the “downstream” side of the shock. The difference between the stagnation temperature is in the moving coordinates

T T = 0 (5.51) 0y − 0x It should be noted that the stagnation temperature (in the stationary coordinates) rises as opposed to the stationary normal shock. The rise in the total temperature is due to the fact that a new material has entered the c.v. at a very high velocity, and is “converted” or added into the total temperature, k 1 2 k 1 2 T T =T 1 + − M M 0 T 1 + − M M 0 0y − 0x y 2 sy − y − x 2 sx − x 0 T0y

k 1 2 k 1 0 = T 1 + − M 0 +T M − (M 2M ) y 2 y y sy 2 sy − y z }| { 0 T0x k 1 2 k 1 T 1 + − M 0 T M − (M 2M ) (5.52) − x 2 x − x sx 2 sx − x z }| { and according to equation (5.51) leads to

Tx k 1 Ty k 1 T 0 T 0 = U − (M 2M ) − (M 2M ) (5.53) 0y − 0x s c 2 sx − x − c 2 sy − y x y Again, this difference in the moving shock is expected because moving material velocity (kinetic energy) is converted into internal energy. This difference can also be viewed as a result of the unsteady state of the shock. 94 CHAPTER 5. NORMAL SHOCK 5.3.1 Shock Result from a Sudden and Complete Stop

The general discussion can be simpliﬁed in the extreme case when the shock is moving from a still medium. This situation arises in many cases in the industry, for example, in a sudden and complete closing of a valve. The sudden closing of the valve must result in a zero velocity of the gas. This shock is viewed by some as a reﬂective shock. The information propagates upstream in which the gas velocity is converted into temperature. In many such cases the steady state is established quite rapidly. In such a case, the shock velocity “downstream” is Us. Equations (5.42) to (5.53) can be transformed into simpler equations when Mx is zero and Us is a positive value. The “upstream” Mach number reads

Us + Ux Mx = = Msx + Mx (5.54) cx

The “downstream” Mach number reads

Us My = | | = Msy (5.55) cy

Again, the shock is moving to the left. In the moving coordinates, the observer (with the shock) sees the ﬂow moving from the left to the right. The ﬂow is moving to the right. The upstream is on the left of the shock. The stagnation temperature increases by

Tx k 1 Ty k 1 T T = U − (M + 2M ) − (M ) (5.56) 0y − 0x s c 2 sx x − c 2 sy

x y ¡£¢j¤¥¡ ¦ The prominent question in this situation is

what will be the shock wave velocity for a 3

0 # ±W²

given ﬂuid velocity, Ux , and for a given W®°¯

5 speciﬁc heat ratio. The “upstream” or the § ¨/©«ª§¬ “downstream” Mach number is not known even if the pressure and the temperature c.v. downstream are given. The difﬁculty lies in Stationary Coordinates the jump from the stationary coordinates to

the moving coordinates. It turns out that it is

À2ÁWÂÃÄÁWÅ:Æ Ç

very useful to use the dimensionless param- ¼W½ ¸ ¹pº¥¸#» eter Msx, or Msy instead of the velocity be- ³3´¶µZ· cause it combines the temperature and the ¾¿ velocity into one parameter. The relationship between the Mach num- c.v. ber on the two sides of the shock are tied Moving Coordinates

Fig. 5.6: Comparison between a stationary shock and a moving shock in a stationary medium in ducts. 5.3. THE MOVING SHOCKS 95 through equations (5.54) and (5.55) by

2 0 2 Mx + Msx + k 1 2 − (My) = 2 (5.57) 2k 0 k 1 Mx + Msx 1 − − And substitutingequation (5.57) into (5.48) results in

f(Msx) 2 0 2 Tx Mx + Msx + k 1 Mx = 2 − (5.58) z }|T { v 2k 0 s y u Mx + M sx 1 u k 1 − t − The temperature ratio in equation Shock in A Suddenly Close Valve (5.58) and the rest of the right–hand k = 1 4 3 side show clearly that Msx has four Msx possible solutions (fourth–order poly- Msy nomial Msx has four solutions). Only one real solution is possible. The so- 2 lution to equation (5.58) can be obtained by several numerical methods. Note, an analytical solution can be ob- 1 tained for equation (5.58) but it seems utilizing numerical methods is much more simple. The typical method is 0 0.1 1 M the “smart” guessing of Msx. For very x small values of the upstream Mach Thu Aug 3 18:54:21 2006 number, M 0 equation (5.58) pro- x ∼ vides that M 1 + 1 and M = Fig. 5.7: The moving shock Mach numbers as a sx ∼ 2 sy 1 1 (the coefficient is only approx- result of a sudden and complete stop. − 2 imated as 0.5) as shown in Figure (5.7). From the same figure it can also be observed that a high velocity can result in a much larger velocity for the reflective shock. For example, a Mach number close to one (1), which can easily be obtained in a Fanno flow, the result is about double the sonic velocity of the reflective shock. Sometimes this phenomenon can have a tremendous significance in industrial applications. Note that to achieve supersonic velocity (in stationary coordinates) a diverging– converging nozzle is required. Here no such device is needed! Luckily and hopefully, engineers who are dealing with a supersonic flow when installing the nozzle and pipe systems for gaseous mediums understand the importance of the reflective shock wave. Two numerical methods and the algorithm employed to solve this problem for given, 0 Mx , is provided herein:

(a) Guess Mx > 1, 96 CHAPTER 5. NORMAL SHOCK

(b) Using shock table or use Potto–GDC to calculate temperature ratio and My,

0 Tx (c) Calculate the Mx = Mx My − Ty q 0 0 (d) Compare to the calculated Mx to the given Mx . and adjust the new guess Mx > 1 accordingly.

The second method is “successive substitutions,” which has better convergence to the solution initially in most ranges but less effective for higher accuracies.

0 (a) Guess Mx = 1 + Mx ,

(b) using the shock table or use Potto–GDC to calculate the temperature ratio and My,

0 Tx (c) calculate the Mx = Mx My − Ty q 0 (d) compare the new Mx approach the old Mx, if not satisfactory use the new Mx to 0 calculate Mx = 1 + Mx then return to part (b).

5.3.2 Moving Shock into Stationary Medium (Suddenly Open Valve)

General Velocities Issues

When a valve or membrane is suddenly opened, a shock is created and propagates downstream. With the exception of close proximity to the valve, the shock moves in a constant velocity (5.8(a)). Using a coordinates system which moves with the shock results in a stationary shock and the ﬂow is moving to the left see Figure

(5.8(b)). The “upstream” will be on the right (see Figure (5.8(b))).

á0âã©á ä

è0éWêÄè ëì è0é6í Î0Ï:Ð¥ÑÓÒÕÔ

×0ØÚÙ Upstream

ÈÊÉ&ËÍÌ

ÝÊÞ&ßÍà

Û#Ü

åçæ Ö

c.v. c.v.

(a) Stationary coordinates (b) Moving coordinates

Fig. 5.8: A shock moves into a still medium as a result of a sudden and complete opening of a valve 5.3. THE MOVING SHOCKS 97 Similar deﬁnitions of the right side and the left side of the shock Mach numbers can be utilized. It has to be noted that the “upstream” and “downstream” are the reverse from the previous case. The “upstream” Mach number is

Us Mx = = Msx (5.59) cx The “downstream” Mach number is

0 Us Uy 0 My = − = Msy My (5.60) cy −

Note that in this case the stagnation temperature in stationary coordinates changes (as in the previous case) whereas the thermal energy (due to pressure difference) is converted into velocity. The stagnation temperature (of moving coordinates) is

k 1 k 1 T T = T 1 + − (M M )2 T 1 + − (M )2 = 0 (5.61) 0y − 0x y 2 sy − y − x 2 x A similar rearrangement to the previous case results in

k 1 2 T 0 T 0 = T 1 + − 2M M + M 2 (5.62) 0y − 0x y 2 − sy y y

Shock in A Suddenly Open Valve Shock in A Suddenly Open Valve k = 1 4, My’ = 0.3 k = 1 4, My’ = 1.3 1.75 4

Mx Mx 3.5 My My T /T T /T 1.5 y x 3 y x

2.5 1.25 2

1.5 1 1

0.5 0.75 0 10 0 5 10 15 20 Number of Iteration Number of Iteration

Wed Aug 23 17:20:59 2006 Wed Aug 23 17:46:15 2006

(a) My 0 = 0.3 (b) My0 = 1.3

Fig. 5.9: The number of iterations to achieve convergence.

The same question that was prominent in the previous case appears now, what will be the shock velocity for a given upstream Mach number? Again, the relationship 98 CHAPTER 5. NORMAL SHOCK between the two sides is

(M )2 + 2 0 sx k 1 M = M + − (5.63) sy y v 2k 2 u k 1 (Msx) 1 u − − t Since Msx can be represented by Msy theoretically equation (5.63) can be solved. It is common practice to solve this equation by numerical methods. One such methods is “successive substitutions.” This method is applied by the following algorithm:

(a) Assume that Mx = 1.0.

(b) Calculate the Mach number My by utilizing the tables or Potto–GDC.

Ty 0 Mx = My + My Tx r calculate the new “improved” Mx.

(d) Check the new and improved Mx against the old one. If it is satisfactory, stop or return to stage (b).

0 To illustrate the convergence of the procedure, consider the case of My = 0.3 and 0 My = 1.3. The results show that the convergence occurs very rapidly (see Figure 0 (5.9)). The larger the value of My , the larger number of the iterations required to achieve the same accuracy. Yet, for most practical purposes, sufﬁcient results can be achieved after 3-4 iterations.

Piston Velocity When a piston is moving, it creates a shock that moves at a speed greater than that of the piston itself. The unknown data are the piston velocity, the temperature, and, other conditions ahead of the shock. Therefore, no Mach number is given but pieces of information on both sides of the shock. In this case, the calculations for Us can be obtained from equation (5.24) that relate the shock velocities and Shock Mach number as

2 Ux Msx (k + 1)Msx = = 2 (5.64) Uy Uy 0 Msx 2 + (k 1)Msx − cx −

Equation (5.64) is a quadratic equation for Msx. There are three solutions of which the ﬁrst one is Msx = 0 and this is immediately disregarded. The other two solutions are

2 0 0 2 (k + 1)Uy Uy (1 + k) + 16cx Msx = (5.65) q4 cx 5.3. THE MOVING SHOCKS 99 The negative sign provides a negative value which is disregarded, and the only solution left is

2 0 0 2 (k + 1)Uy + Uy (1 + k) + 16cx Msx = (5.66) q4 cx or in a dimensionless form

2 0 0 (k + 1)Myx + Myx (1 + k) + 16 Msx = (5.67) q4

0 0 Where the “stange” Mach number is Myx = Uy /cx. The limit of the equation when c leads to x → ∞ (k + 1)M 0 M = yx (5.68) sx 4 As one additional “strange” it can be seen that the shock is close to the piston when the gas ahead of the piston is very hot. This phenomenon occurs in many industrial applications, such as the internal combustion engines and die casting. Some use equation (5.68) to explain the next Shock-Choke phenomenon.

Shock–Choke Phenomenon Assuming that the gas velocity is supersonic (in stationary coordinates) before the shock moves, what is the maximum velocity that can be reached before this model fails? In other words, is there a point where the moving shock is fast enough to reduce the “upstream” relative Mach number below the speed of sound? This is the point where regardless of the pressure difference is, the shock Mach number cannot be increased. This shock–choking phenomenon Shock in A Suddenly Open Valve Maximum My’ possible is somewhat similar to the 2.5

My choking phenomenon that was 2.25 (max) discussed earlier in a nozzle 2 ’ ﬂow and in other pipe ﬂow mod- y 1.75 els (later chapters). The differ- 1.5

ence is that the actual velocity Maximum M 1.25 has no limit. It must be noted 1 that in the previous case of 0.75 suddenly and completely clos- 0.5 The spesific heat ratio, k ing of valve results in no limit

(at least from the model point Thu Aug 24 17:46:07 2006 of view). To explain this phenomenon, look at the normal Fig. 5.10: The maximum of “downstream” Mach num- k shock. Consider when the “up- ber as a function of the speciﬁc heat, . stream” Mach approaches inﬁnity, M = M , and the downstream Mach x sx → ∞ 100 CHAPTER 5. NORMAL SHOCK number, according to equation (5.38), is approaching to (k 1)/2k. One can view − this as the source of the shock–choking phenomenon. These limits determine the maximum velocity after the shock since Umax = cyMy. From the upstream side, the Mach number is

∞ T k 1 M = M = y − (5.69) x sx T 2k r x Thus, the Mach number is approaching infinity because of the temperature ratio but the velocity is finite. To understand this limit, consider that the maximum Mach number is obtained when the pressure ratio is approaching infinity Py . By applying equation Px → ∞ (5.23) to this situation the following is obtained:

k + 1 Px M = 1 + 1 (5.70) sx 2k P − s y and the mass conservation leads to

Uyρy = Usρx U U 0 ρ = U ρ s − y y s x T ρ M 0 = y 1 x M (5.71) y T − ρ sx r x y Substituting equations (5.26) and (5.25) into equation (5.71) results in

2k 1 + k+1 Py 1 P k 1 Px 0 y k+1 − My = 1 (5.72) Py 1 v k − Px v k × k+1 Py P + k+1− u + u x u k 1 Px u u − t t When the pressure ratio is approaching inﬁnity (extremely strong pressure ratio), the results is

2 M 0 = (5.73) y k(k 1) s −

What happens when a gas with a Mach number larger than the maximum Mach number possible is flowing in the tube? Obviously, the semi steady state described by the moving shock cannot be sustained. A similar phenomenon to the choking in the nozzle and later in an internal pipe flow is obtained. The Mach number is reduced to the maximum value very rapidly. The reduction occurs by an increase of temperature after the shock or a stationary shock occurs as it will be shown in chapters on internal flow. 5.3. THE MOVING SHOCKS 101

0 Ty k Mx My My Tx 1.30 1073.25 0.33968 2.2645 169842.29 1.40 985.85 0.37797 1.8898 188982.96 1.50 922.23 0.40825 1.6330 204124.86 1.60 873.09 0.43301 1.4434 216507.05 1.70 833.61 0.45374 1.2964 226871.99 1.80 801.02 0.47141 1.1785 235702.93 1.90 773.54 0.48667 1.0815 243332.79 2.00 750.00 0.50000 1.00000 250000.64 2.10 729.56 0.51177 0.93048 255883.78 2.20 711.62 0.52223 0.87039 261117.09 2.30 695.74 0.53161 0.81786 265805.36 2.40 681.56 0.54006 0.77151 270031.44 2.50 668.81 0.54772 0.73029 273861.85

Table of maximum values of the shock-choking phenomenon. The mass ﬂow rate when the pressure ratio is approaching inﬁnity, , is ∞

ρy cy

m˙ 0 0 0 Py = Uy ρy = My cyρy = My kRTy A zR}|Ty{ pz }| { M 0 √kP = y y (5.74) RTy

Equation (5.74) and equation (5.25)pcan be transferred for large pressure ratios into

m˙ Px k 1 Ty − (5.75) A ∼ Tx k + 1 p

Since the right hand side of equation (5.75) is constant, with the exception of Ty the mass flow rate is approaching infinity when the pressure ratio is approaching p infinity. Thus, the shock–choke phenomenon means that the Mach number is only limited in stationary coordinates but the actual flow rate isn’t.

5.3.3 Partially Open Valve The previous case is a special case of the moving shock. The general case is when one gas ﬂows into another gas with a given velocity. The only limitation is that the “downstream’ gas velocity is higher than the “upstream” gas velocity as shown in Figure (5.13). 102 CHAPTER 5. NORMAL SHOCK

¡ £¢Äÿ¥¤§¦ ÿ¡ ©¨ 0 ÿ U = U − U ò x s x

ô0õÚö Ux Upstream îÊï&ðÍñ

0 0 ùÊú&ûÍü

÷#ø ýçþ

ó Uy > Ux

c.v. c.v.

(a) Stationary coordinates (b) Moving coordinates

Fig. 5.11: A shock moves into a moving medium as a result of a sudden and complete open valve.

The relationship between the different Mach numbers on the “upstream” side is

M = M M 0 (5.76) x sx − x The relationship between the different Mach on the “downstream” side is

M = M M 0 (5.77) y sy − y An additional parameter has be supplied to solve the problem. A common problem is to ﬁnd the moving shock velocity when the velocity “downstream” or the pressure is suddenly increased. It has to be mentioned that the temperature “downstream” is unknown (the ﬂow of the gas with the higher velocity). The procedure for the calculations can be done by the following algorithm:

0 (a) Assume that Mx = Mx + 1.

(b) Calculate the Mach number My by utilizing the tables or Potto–GDC.

0 (c) Calculate the “downstream” shock Mach number Msy = My + My

(d) Utilizing T M = y (M ) M 0 x T sy x r x −

calculate the new “improved” Mx

(e) Check the new and improved Mx against the old one. If it is satisfactory, stop or return to stage (b). 5.3. THE MOVING SHOCKS 103

Earlier, it was shown that the shock chok- Shock in A Suddenly Open Valve k = 1 4 ing phenomenon occurs when the flow is 1 running into a still medium. This phe- 0.9 nomenon also occurs in the case where M ’ = 0.9 0.8 x Mx’ = 0.2 a faster flow is running into a slower fluid. M ’ = 0.0 0.7 x y

The mathematics is cumbersome but re- M sults show that the shock choking phe- 0.6 nomenon is still there (the Mach number is 0.5 limited, not the actual ﬂow). Figure (5.12) 0.4

0.3 exhibits some “downstream” Mach num- 0.4 0.8 1.2 1.6 2 2.4 2.8 0 M ’ bers for various static Mach numbers, My , y and for various static “upstream” Mach Thu Oct 19 10:34:19 2006 numbers, M 0 . The ﬁgure demonstrates x Fig. 5.12: The results of the partial opening that the maximum can also occurs in the of the valve. vicinity of the previous value (see following question/example).

5.3.4 Partially Closed Valve

0 0 0 0 U = U + U Uy = Us + Uy U U x s x x y Upstream

ρy P y Ty

c.v. c.v.

(a) Stationary coordinates (b) Moving coordinates

Fig. 5.13: A shock as a result of a sudden and partially a valve closing or a narrowing the passage to the ﬂow

The totally closed valve is a special case of a partially closed valve in which there is a sudden change and the resistance increases in the pipe. The information propagates upstream in the same way as before. Similar equations can be written:

0 Ux = Us + Ux (5.78)

0 Uy = Us + Uy (5.79) 104 CHAPTER 5. NORMAL SHOCK

0 Mx = Ms + Mx (5.80)

0 My = Ms + My (5.81)

For given static Mach numbers the procedure for the calculation is as follows:

0 (a) Assume that Mx = Mx + 1.

(b) . Calculate the Mach number My by utilizing the tables or Potto–GDC

(c) Calculate the “downstream” shock Mach number M = M M 0 sy y − y (d) Utilizing T M = y (M ) + M 0 x T sy x r x calculate the new “improved” Mx

(e) Check the new and improved Mx against the old one. If it is satisfactory, stop or return to stage (b).

5.3.5 Worked–out Examples for Shock Dynamics Example 5.2: A shock is moving at a speed of 450 [m/sec] in a stagnated gas at pressure of 1 [Bar] and temperature of 27◦C. Compute the pressure and the temperature behind the shock. Assume the speciﬁc heat ratio is 1.3.

SOLUTION It can be observed that the gas behind the shock is moving while the gas ahead of the shock is still. Thus, it is the case of a shock moving into still medium (suddenly opened valve case). First, the Mach velocity ahead of the shock has to calculated.

0 U 450 My = = 1.296 √kRT √1.3 287 300 ∼ × × By utilizing Potto–GDC or Table (5.4) one can obtain the following table:

0 0 Ty Py P0y Mx My Mx My Tx Px P0x 2.1206 0.54220 0.0 1.132 1.604 4.953 0.63955 5.3. THE MOVING SHOCKS 105 Using the above table, the temperature behind the shock is

0 Ty Ty = Ty = Tx = 1.604 300 481.2K Tx × ∼ In same manner, it can be done for the pressure ratio as following

0 Py Py = Py = Px = 4.953 1.0 4.953[Bar] Px × ∼ The velocity behind the shock wave is obtained m U 0 = M 0 c = 1.132 √1.3 287 300 378.72 y x x × × × ∼ sec h i

Example 5.3: Gas flows in a tube with a velocity of 450[m/sec]. The static pressure at the tube is 2Bar and the (static) temperature of 300K. The gas is brought into a complete stop by a sudden closing a valve. Calculate the velocity and the pressure behind the reflecting shock. The specific heat ratio can be assumed to be k = 1.4.

SOLUTION 0 The first thing that needs to be done is to find the prime Mach number Mx = 1.2961. Then, the prime properties can be found. At this stage the reflecting shock velocity is unknown. Simply using the Potto–GDC provides for the temperature and velocity the following table:

0 0 Ty Py P0y Mx My Mx My Tx Px P0x 2.0445 0.56995 1.2961 0.0 1.724 4.710 0.70009

0 If you insist on doing the steps yourself, find the upstream prime Mach, Mx to be 1.2961. Then using Table (5.2) you can find the proper Mx. If this detail is not sufficient then simply utilize the iterations procedure described earlier and obtain the following:

Ty 0 i Mx My My Tx 0 2.2961 0.53487 1.9432 0.0 1 2.042 0.57040 1.722 0.0 2 2.045 0.56994 1.724 0.0 3 2.044 0.56995 1.724 0.0 4 2.044 0.56995 1.724 0.0

The table was obtained by utilizing Potto–GDC with the iteration request. 106 CHAPTER 5. NORMAL SHOCK Example 5.4: What should be the prime Mach number (or the combination of the velocity with the temperature, for those who like an additional step) in order to double the temperature when the valve is suddenly and totally closed?

SOLUTION The ratio can be obtained from Table (5.3). It can also be obtained from the stationary normal shock wave table. Potto-GDC provides for this temperature ratio the following table:

Ty ρy Py P0y Mx My Tx ρx Px P0x 2.3574 0.52778 2.0000 3.1583 6.3166 0.55832

using the required Mx = 2.3574 in the moving shock table provides

0 0 Ty Py P0y Mx My Mx My Tx Px P0x 2.3574 0.52778 0.78928 0.0 2.000 6.317 0.55830

Example 5.5: A gas is ﬂowing in a pipe with a Mach number of 0.4. Calculate the speed of the shock when a valve is closed in such a way that the Mach number is reduced by half. Hint, this is the case of a partially closed valve case in which the ratio of the prime Mach number is half (the new parameter that is added in the general case).

SOLUTION Refer to section (5.3.4) for the calculation procedure. Potto-GDC provides the solution of the above data

0 0 Ty Py P0y Mx My Mx My Tx Px P0x 1.1220 0.89509 0.40000 0.20000 1.0789 1.3020 0.99813

If the information about the iterations is needed please refer to the following table. 5.3. THE MOVING SHOCKS 107

Ty Py 0 i Mx My My Tx Px 0 1.4000 0.73971 1.2547 2.1200 0.20000 1 1.0045 0.99548 1.0030 1.0106 0.20000 2 1.1967 0.84424 1.1259 1.5041 0.20000 3 1.0836 0.92479 1.0545 1.2032 0.20000 4 1.1443 0.87903 1.0930 1.3609 0.20000 5 1.1099 0.90416 1.0712 1.2705 0.20000 6 1.1288 0.89009 1.0832 1.3199 0.20000 7 1.1182 0.89789 1.0765 1.2922 0.20000 8 1.1241 0.89354 1.0802 1.3075 0.20000 9 1.1208 0.89595 1.0782 1.2989 0.20000 10 1.1226 0.89461 1.0793 1.3037 0.20000 11 1.1216 0.89536 1.0787 1.3011 0.20000 12 1.1222 0.89494 1.0790 1.3025 0.20000 13 1.1219 0.89517 1.0788 1.3017 0.20000 14 1.1221 0.89504 1.0789 1.3022 0.20000 15 1.1220 0.89512 1.0789 1.3019 0.20000 16 1.1220 0.89508 1.0789 1.3020 0.20000 17 1.1220 0.89510 1.0789 1.3020 0.20000 18 1.1220 0.89509 1.0789 1.3020 0.20000 19 1.1220 0.89509 1.0789 1.3020 0.20000 20 1.1220 0.89509 1.0789 1.3020 0.20000 21 1.1220 0.89509 1.0789 1.3020 0.20000 22 1.1220 0.89509 1.0789 1.3020 0.20000

Example 5.6: A piston is pushing air that ﬂows in a tube with a Mach number of M = 0.4 and 300◦C. The piston is 0 0 Mx = 0.4 accelerated very rapidly and the air My = 0.8 adjoined the piston obtains Mach number M = 0.8. Calculate the velocity of the shock created by the piston in the air. Calculate the time Fig. 5.14: Schematic of a piston pushing air in a it takes for the shock to reach the tube. end of the tube of 1.0m length. Assume that there is no friction and the Fanno ﬂow model is not applicable.

SOLUTION Using the procedure described in this section, the solution is 108 CHAPTER 5. NORMAL SHOCK

0 0 Ty Py P0y Mx My Mx My Tx Px P0x 1.2380 0.81942 0.50000 0.80000 1.1519 1.6215 0.98860

The complete iteration is provided below.

Ty Py 0 i Mx My My Tx Px 0 1.5000 0.70109 1.3202 2.4583 0.80000 1 1.2248 0.82716 1.1435 1.5834 0.80000 2 1.2400 0.81829 1.1531 1.6273 0.80000 3 1.2378 0.81958 1.1517 1.6207 0.80000 4 1.2381 0.81940 1.1519 1.6217 0.80000 5 1.2380 0.81943 1.1519 1.6215 0.80000 6 1.2380 0.81942 1.1519 1.6216 0.80000

The time it takes for the shock to reach the end of the cylinder is

length 1 t = = = 0.0034[sec] Us √1.4 287 300(1.2380 0.4) × × − c (M M 0 ) x x− x |{z}

Example 5.7: From the previous example (5.10) calculate the velocity difference between initial piston velocity and ﬁnal piston velocity.

SOLUTION The stationary difference between the two sides of the shock is:

∆U =U 0 U 0 = c U 0 c U 0 y − x y y − x x

Ty Tx q = 1.4 287 300 0.8 √1.1519 0.5 × × × − q  z }| {      124.4[m/sec]   ∼ 5.4. SHOCK TUBE 109 Example 5.8: An engine is designed so that two pistons are moving toward each other (see Figure (5.15)). 1 [Bar] The air between the pistons is 300 K at 1[Bar] and 300K. The dis- 40 m/sec 70 m/sec shock tance between the two pistons waves is 1[m]. Calculate the time it will take for the two shocks to collide. Fig. 5.15: Figure for Example (5.8) SOLUTION This situation is an open valve case where the prime information is given. The solution is given by equation (5.66), and, it is the explicit analytical solution. For this case the following table can easily be obtain from Potto–GDC for the left piston

0 0 Ty Py P0y 0 Mx My Mx My Uy cx Tx Px P0x 1.0715 0.93471 0.0 0.95890 1.047 1.173 0.99959 40.0 347. while the velocity of the right piston is

0 0 Ty Py P0y 0 Mx My Mx My Uy cx Tx Px P0x 1.1283 0.89048 0.0 0.93451 1.083 1.318 0.99785 70.0 347.

The time for the shocks to collide is length 1[m] t = = 0.0013[sec] Usx1 + Usx2 (1.0715 + 1.1283)347. ∼

5.4 Shock Tube The shock tube is a study tool with very little practical purposes. It is used in many cases to understand certain phenomena. Other situations can be examined and extended from these phenomena. A cylinder with two chambers connected by a diaphragm. On one side the pressure is high, while the pressure on the other side is low. When the diaphragm is ruptured the gas from the high pressure section flows into the low pressure section. When the pressure is high enough, a shock is created that it travels to the low pressure chamber. This is the same case as in the suddenly opened valve case described previously. At the back of the shock, expansion waves occur with a reduction of pressure. The temperature is known to reach several thousands degrees in a very brief period of time. The high pressure 110 CHAPTER 5. NORMAL SHOCK chamber is referred to in the literature is the driver section and the low section is referred to as the expansion section. Initially, the gas from the driver section is coalescing from small shock waves into a large shock wave. In this analysis, it is assumed that this time is essentially zero. Zone 1 is an undisturbed gas and zone 2 is an area where the shock already passed. The assumption is that the shock is very sharp with zero width. On the other side, the expansion waves are moving into the high pressure chamber i.e. the driver section. The shock is moving at a supersonic speed (it depends on the definition, i.e., what reference temperature is being used) and the medium behind the shock is also moving but at a velocity, U2, which can be supersonic or subsonic in stationary coordinates. The velocities in the expansion chamber vary between three zones. In zone 3 is the original material that was in the high pressure chamber but is now the same pressure as zone 2. Zone 4 is where the gradual transition occurs between original high pressure to low pressure. The boundaries of zone 4 are defined by initial conditions. The expansion front is moving at the local speed of sound in the high pressure section. The expansion back front is moving at the local speed of sound velocity but the actual gas is moving in the opposite direction in U2. In fact, material in the expansion chamber and the front are moving to the left while the actual flow of the gas is moving to the right (refer to Figure (5.16)). In zone 5, the velocity is zero and the pressure is in its original value. The properties in the 5 expansion 4 3 2 1 different zones have front different relationships. Diaphragm The relationship be- t tween zone 1 and zone 2 is that of a moving shock into still medium some where reflective reflective wave shock (again, this is a case of wave sudden opened valve). The material in zone t1 back 2 and 3 is moving front at the same velocity Contact Surface shock wave (speed) but the temperature and the entropy are different, while the distance pressure in the two zones are the same. Fig. 5.16: The shock tube schematic with a pressure ”diagram.” The pressure, the temperature and their properties in zone 4 aren’t constant and continuous between the conditions in zone 3 to the conditions in zone 5. The expansion front wave velocity is larger than the velocity at the back front expansion wave velocity. Zone 4 is expanding during the initial stage (until the expansion reaches the wall). The shock tube is a relatively small length 1 2[m] and the typical velocity is in the − 5.4. SHOCK TUBE 111 range of the speed of sound, c √340 thus the whole process takes only a few ∼ milliseconds or less. Thus, these kinds of experiments require fast recording devices (a relatively fast camera and fast data acquisition devices.). A typical design problem of a shock tube is finding the pressure to achieve the desired temperature or Mach number. The relationship between the different properties was discussed earlier and because it is a common problem, a review of the material is provided thus far. The following equations were developed earlier and are repeated here for clarifica- tion. The pressure ratio between the two sides of the shock is P k 1 2k 2 = − M 2 1 (5.82) P k + 1 k 1 s1 − 1 − Rearranging equation (5.82) becomes

k 1 k + 1 P M = − + 2 (5.83) s1 2k 2k P r 1 Or expressing the velocity as

k 1 k + 1 P2 Us = Ms1c1 = c1 − + (5.84) r 2k 2k P1 And the velocity ratio between the two sides of the shock is

k+1 P2 U1 ρ2 1 + k 1 P = = − 1 (5.85) U ρ k+1 P2 2 2 k 1 P − 1 The ﬂuid velocity in zone 2 is the same

U2 U 0 = U U = U 1 (5.86) 2 s − 2 s − U s From the mass conservation, it follows that U ρ 2 = 1 (5.87) Us ρ2

k+1 P2 k 1 k + 1 P2 k 1 + P 0 − 1 U2 = c1 − + 1 k+1 P (5.88) 2k 2k P1 v − 1 + 2 r u k 1 P1 u − t After rearranging equation (5.88) the result is

2k 0 c1 P2 k+1 U2 = 1 (5.89) P2 k 1 k P1 − v − u P1 1+k u t 112 CHAPTER 5. NORMAL SHOCK

On the isentropic side, in zone 4, taking the derivative of the continuity equation, d(ρU) = 0, and dividing by the continuity equation the following is obtained: dρ dU = (5.90) ρ − c Since the process in zone 4 is isentropic, applying the isentropic relationship (T k 1 ∝ ρ − ) yields

k 1 c T ρ −2 = = (5.91) c T ρ 5 r 5 5 From equation (5.90) it follows that

k 1 dρ ρ −2 dU = c = c dρ (5.92) − ρ 5 ρ 5 Equation (5.92) can be integrated as follows:

k 1 U3 ρ3 ρ −2 dU = c dρ (5.93) 5 ρ ZU5=0 Zρ5 5 The results of the integration are

k 1 2c ρ −2 U = 5 1 3 (5.94) 3 k 1 − ρ − 5 ! Or in terms of the pressure ratio as

k 1 2c P 2−k U = 5 1 3 (5.95) 3 k 1 − P − 5 ! As it was mentioned earlier the velocity at points 20 and 3 are identical, hence equation (5.95) and equation (5.89) can be combined to yield

k 1 − 2k 2c P 2k c P 5 1 3 = 1 2 1 k+1 (5.96) P2 k 1 k 1 − P5 k P1 − v − ! u P1 1+k − u t After some rearrangement, equation (5.96) is transformed into

2k k 1 − − (k 1) c1 P5 1 P5 P2 c5 P3 = 1 − −  (5.97) P1 P1 − √2k 2k + (k + 1) P2 1  P1   −   r  5.5. SHOCK WITH REAL GASES 113

Or in terms of the Mach number, Ms1

2k k 1 c1 2 k 1 P k 1 2k − Ms1 1 − − 5 = 1 − M 2 1 1 k+1 c5 − (5.98) P k + 1 + 1 k 1 s1 − − M 1 1 − " s1 #

Using the Rankine–Hugoniot relationship and the perfect gas model, the following is obtained:

k 1 P 1 + 1− 2 T2 k1+1 P1 = k 1 P (5.99) T1 1 + 1− 1 k1+1 P2

By utilizing the isentropic relationship for zone 3 to 5 results in

k5 1 k5 1 − − P2 k5 k5 T3 P3 P1 = = P (5.100) T5 P5 5 P1 !

Example 5.9: A shock tube with an initial pressure ratio of P5 = 20 and an initial temperature of P1 300K. Find the shock velocity and temperature behind the shock if the pressure ratio is P5 = 40? P1

SOLUTION

5.5 Shock with Real Gases

5.6 Shock in Wet Steam

5.7 Normal Shock in Ducts

The flow in ducts is related to boundary layer issues. For a high Reynolds number, the assumption of an uniform flow in the duct is closer to reality. It is normal to have a large Mach number with a large Re number. In that case, the assumptions in construction of these models are acceptable and reasonable. 114 CHAPTER 5. NORMAL SHOCK 5.8 More Examples for Moving Shocks Example 5.10: This problem was taken from the real industrial manufactur- distance ing world. An engineer is required to design a cooling system for a critical electronic device. The temperature should valve exit not increase above a certain value. In this system, air is Fig. 5.17: Figure for Example (5.10) supposed to reach the pipe exit as quickly as possible when the valve is opened (see Figure (5.17)). opening valve probelm The distance between between the valve and the pipe exit is 3[m]. The conditions upstream of the valve are 30[Bar] and 27◦C . Assume that there isn’t any resistance whatsoever in the pipe. The ambient temperature is 27◦C and 1[Bar]. Assume that the time scale for opening the valve is significantly smaller than the typical time of the pipe (totally unrealistic even though the valve manufacture claims of 0.0002 [sec] to be opened). After building the system, the engineer notices that the system does not cool the device fast enough and proposes to increase the pressure and increase the diameter of the pipe. Comment on this proposal. Where any of these advises make any sense in the light of the above assumptions? What will be your recommendations to the manufacturing company? Plot the exit temperature and the mass flow rate as a function of the time.

SOLUTION This problem is known as the suddenly open valve problem in which the shock choking phenomenon occurs. The time it takes for the shock to travel from the valve depends on the pressure ratio Py = 30 Px

0 0 Ty Py P0y Mx My Mx My Tx Px P0x 5.0850 0.41404 0.0 1.668 5.967 30.00 0.057811

The direct calculation will be by using the “upstream” Mach number, Mx = Msx = 5.0850. Therefore, the time is distance 3 t = = 0.0017[sec] M √kRT 5.0850sqrt1.4 287 300 ∼ sx x × × The mass ﬂow rate after reaching the exit under these assumptions remains constant until the uncooled material reaches the exit. The time it takes for the material from the valve to reach the exit is distance 3 t = = 0.0021[sec] M 0 kRT 1.668sqrt1.4 287 300 5.967 ∼ y y × × × p 5.9. TABLES OF NORMAL SHOCKS, K = 1.4 IDEAL GAS 115

During that difference of time the material is get heated instead of cooling down because of the high temperature. The suggestion of the engineer to in- Mass Flow crease the pressure will decrease the Rate time but will increase the temperature at the exit during this critical time period. Thus, this suggestion contradicts Velocity the purpose of the required manufacturing needs. To increase the pipe diameter will not Time[Msec] change the temperature and therefore will not change the effects of heating. It Fig. 5.18: The results for Example (5.10) can only increase the rate after the initial heating spike A possible solution is to have the valve very close to the pipe exit. Thus, the heating time is reduced signiﬁcantly. There is also the possibility of steps increase in which every step heat released will not be enough to over heat the device. The last possible requirement a programmable valve and very fast which its valve probably exceed the moving shock the valve downstream. The plot of the mass ﬂow rate and the velocity are given in Figure (5.18).

Example 5.11: Example (5.10) deals with a damaging of electronic product by the temperature increase. Try to estimate the temperature increase of the product. Plot the pipe exit temperature as a function of the time.

SOLUTION

5.9 Tables of Normal Shocks, k = 1.4 Ideal Gas

Table 5.1: The shock wave table for k = 1.4

Ty ρy Py P0y Mx My Tx ρx Px P0x 1.00 1.00000 1.00000 1.00000 1.00000 1.00000 1.05 0.95313 1.03284 1.08398 1.11958 0.99985 1.10 0.91177 1.06494 1.16908 1.24500 0.99893 1.15 0.87502 1.09658 1.25504 1.37625 0.99669 1.20 0.84217 1.12799 1.34161 1.51333 0.99280 1.25 0.81264 1.15938 1.42857 1.65625 0.98706 1.30 0.78596 1.19087 1.51570 1.80500 0.97937 116 CHAPTER 5. NORMAL SHOCK Table 5.1: The shock wave table for k = 1.4 (continue)

Ty ρy Py P0y Mx My Tx ρx Px P0x 1.35 0.76175 1.22261 1.60278 1.95958 0.96974 1.40 0.73971 1.25469 1.68966 2.12000 0.95819 1.45 0.71956 1.28720 1.77614 2.28625 0.94484 1.50 0.70109 1.32022 1.86207 2.45833 0.92979 1.55 0.68410 1.35379 1.94732 2.63625 0.91319 1.60 0.66844 1.38797 2.03175 2.82000 0.89520 1.65 0.65396 1.42280 2.11525 3.00958 0.87599 1.70 0.64054 1.45833 2.19772 3.20500 0.85572 1.75 0.62809 1.49458 2.27907 3.40625 0.83457 1.80 0.61650 1.53158 2.35922 3.61333 0.81268 1.85 0.60570 1.56935 2.43811 3.82625 0.79023 1.90 0.59562 1.60792 2.51568 4.04500 0.76736 1.95 0.58618 1.64729 2.59188 4.26958 0.74420 2.00 0.57735 1.68750 2.66667 4.50000 0.72087 2.05 0.56906 1.72855 2.74002 4.73625 0.69751 2.10 0.56128 1.77045 2.81190 4.97833 0.67420 2.15 0.55395 1.81322 2.88231 5.22625 0.65105 2.20 0.54706 1.85686 2.95122 5.48000 0.62814 2.25 0.54055 1.90138 3.01863 5.73958 0.60553 2.30 0.53441 1.94680 3.08455 6.00500 0.58329 2.35 0.52861 1.99311 3.14897 6.27625 0.56148 2.40 0.52312 2.04033 3.21190 6.55333 0.54014 2.45 0.51792 2.08846 3.27335 6.83625 0.51931 2.50 0.51299 2.13750 3.33333 7.12500 0.49901 2.75 0.49181 2.39657 3.61194 8.65625 0.40623 3.00 0.47519 2.67901 3.85714 10.33333 0.32834 3.25 0.46192 2.98511 4.07229 12.15625 0.26451 3.50 0.45115 3.31505 4.26087 14.12500 0.21295 3.75 0.44231 3.66894 4.42623 16.23958 0.17166 4.00 0.43496 4.04688 4.57143 18.50000 0.13876 4.25 0.42878 4.44891 4.69919 20.90625 0.11256 4.50 0.42355 4.87509 4.81188 23.45833 0.09170 4.75 0.41908 5.32544 4.91156 26.15625 0.07505 5.00 0.41523 5.80000 5.00000 29.00000 0.06172 5.25 0.41189 6.29878 5.07869 31.98958 0.05100 5.50 0.40897 6.82180 5.14894 35.12500 0.04236 5.75 0.40642 7.36906 5.21182 38.40625 0.03536 6.00 0.40416 7.94059 5.26829 41.83333 0.02965 6.25 0.40216 8.53637 5.31915 45.40625 0.02498 5.9. TABLES OF NORMAL SHOCKS, K = 1.4 IDEAL GAS 117 Table 5.1: The shock wave table for k = 1.4 (continue)

Ty ρy Py P0y Mx My Tx ρx Px P0x 6.50 0.40038 9.15643 5.36508 49.12500 0.02115 6.75 0.39879 9.80077 5.40667 52.98958 0.01798 7.00 0.39736 10.46939 5.44444 57.00000 0.01535 7.25 0.39607 11.16229 5.47883 61.15625 0.01316 7.50 0.39491 11.87948 5.51020 65.45833 0.01133 7.75 0.39385 12.62095 5.53890 69.90625 0.00979 8.00 0.39289 13.38672 5.56522 74.50000 0.00849 8.25 0.39201 14.17678 5.58939 79.23958 0.00739 8.50 0.39121 14.99113 5.61165 84.12500 0.00645 8.75 0.39048 15.82978 5.63218 89.15625 0.00565 9.00 0.38980 16.69273 5.65116 94.33333 0.00496 9.25 0.38918 17.57997 5.66874 99.65625 0.00437 9.50 0.38860 18.49152 5.68504 105.12500 0.00387 9.75 0.38807 19.42736 5.70019 110.73958 0.00343 10.00 0.38758 20.38750 5.71429 116.50000 0.00304

Table 5.2: Table for a Reﬂective Shock from a suddenly closed end (k=1.4)

0 0 Ty Py P0y Mx My Mx My Tx Px P0x 1.006 0.99403 0.01 0.0 1.004 1.014 1.00000 1.012 0.98812 0.02 0.0 1.008 1.028 1.00000 1.018 0.98227 0.03 0.0 1.012 1.043 0.99999 1.024 0.97647 0.04 0.0 1.016 1.057 0.99998 1.030 0.97074 0.05 0.0 1.020 1.072 0.99997 1.037 0.96506 0.06 0.0 1.024 1.087 0.99994 1.043 0.95944 0.07 0.0 1.028 1.102 0.99991 1.049 0.95387 0.08 0.0 1.032 1.118 0.99986 1.055 0.94836 0.09 0.0 1.036 1.133 0.99980 1.062 0.94291 0.10 0.0 1.040 1.149 0.99973 1.127 0.89128 0.20 0.0 1.082 1.316 0.99790 1.196 0.84463 0.30 0.0 1.126 1.502 0.99317 1.268 0.80251 0.40 0.0 1.171 1.710 0.98446 1.344 0.76452 0.50 0.0 1.219 1.941 0.97099 1.423 0.73029 0.60 0.0 1.269 2.195 0.95231 1.505 0.69946 0.70 0.0 1.323 2.475 0.92832 1.589 0.67171 0.80 0.0 1.381 2.780 0.89918 1.676 0.64673 0.90 0.0 1.442 3.112 0.86537 1.766 0.62425 1.00 0.0 1.506 3.473 0.82755 118 CHAPTER 5. NORMAL SHOCK Table 5.2: Table for Reﬂective Shock from suddenly closed valve (end) (k=1.4)(continue)

0 0 Ty Py P0y Mx My Mx My Tx Px P0x 1.858 0.60401 1.10 0.0 1.576 3.862 0.78652 1.952 0.58578 1.20 0.0 1.649 4.280 0.74316 2.048 0.56935 1.30 0.0 1.727 4.728 0.69834 2.146 0.55453 1.40 0.0 1.810 5.206 0.65290 2.245 0.54114 1.50 0.0 1.897 5.715 0.60761 2.346 0.52904 1.60 0.0 1.990 6.256 0.56312 2.448 0.51808 1.70 0.0 2.087 6.827 0.51996 2.552 0.50814 1.80 0.0 2.189 7.431 0.47855 2.656 0.49912 1.90 0.0 2.297 8.066 0.43921 2.762 0.49092 2.00 0.0 2.410 8.734 0.40213 3.859 0.43894 3.00 0.0 3.831 17.21 0.15637 5.000 0.41523 4.00 0.0 5.800 29.00 0.061716 6.162 0.40284 5.00 0.0 8.325 44.14 0.026517 7.336 0.39566 6.00 0.0 11.41 62.62 0.012492 8.517 0.39116 7.00 0.0 15.05 84.47 0.00639 9.703 0.38817 8.00 0.0 19.25 1.1E+2 0.00350 10.89 0.38608 9.00 0.0 24.01 1.4E+2 0.00204 12.08 0.38457 10.0 0.0 29.33 1.7E+2 0.00125

Table 5.3: Table for shock propagating from suddenly opened valve (k=1.4)

0 0 Ty Py P0y Mx My Mx My Tx Px P0 x 1.006 0.99402 0.0 0.01 1.004 1.014 1.00000 1.012 0.98807 0.0 0.02 1.008 1.028 1.00000 1.018 0.98216 0.0 0.03 1.012 1.043 0.99999 1.024 0.97629 0.0 0.04 1.016 1.058 0.99998 1.031 0.97045 0.0 0.05 1.020 1.073 0.99996 1.037 0.96465 0.0 0.06 1.024 1.088 0.99994 1.044 0.95888 0.0 0.07 1.029 1.104 0.99990 1.050 0.95315 0.0 0.08 1.033 1.120 0.99985 1.057 0.94746 0.0 0.09 1.037 1.136 0.99979 1.063 0.94180 0.0 0.10 1.041 1.152 0.99971 1.133 0.88717 0.0 0.20 1.086 1.331 0.99763 1.210 0.83607 0.0 0.30 1.134 1.541 0.99181 1.295 0.78840 0.0 0.40 1.188 1.791 0.98019 1.390 0.74403 0.0 0.50 1.248 2.087 0.96069 1.495 0.70283 0.0 0.60 1.317 2.441 0.93133 1.613 0.66462 0.0 0.70 1.397 2.868 0.89039 5.9. TABLES OF NORMAL SHOCKS, K = 1.4 IDEAL GAS 119 Table 5.3: Table for shock propagating from suddenly opened valve (k=1.4)

0 0 Ty Py P0y Mx My Mx My Tx Px P0x 1.745 0.62923 0.0 0.80 1.491 3.387 0.83661 1.896 0.59649 0.0 0.90 1.604 4.025 0.76940 2.068 0.56619 0.0 1.00 1.744 4.823 0.68907 2.269 0.53817 0.0 1.100 1.919 5.840 0.59699 2.508 0.51223 0.0 1.200 2.145 7.171 0.49586 2.799 0.48823 0.0 1.300 2.450 8.975 0.38974 3.167 0.46599 0.0 1.400 2.881 11.54 0.28412 3.658 0.44536 0.0 1.500 3.536 15.45 0.18575 4.368 0.42622 0.0 1.600 4.646 22.09 0.10216 5.551 0.40843 0.0 1.700 6.931 35.78 0.040812 8.293 0.39187 0.0 1.800 14.32 80.07 0.00721 8.821 0.39028 0.0 1.810 16.07 90.61 0.00544 9.457 0.38870 0.0 1.820 18.33 1.0E + 2 0.00395 10.24 0.38713 0.0 1.830 21.35 1.2E + 2 0.00272 11.25 0.38557 0.0 1.840 25.57 1.5E + 2 0.00175 12.62 0.38402 0.0 1.850 31.92 1.9E + 2 0.00101 14.62 0.38248 0.0 1.860 42.53 2.5E + 2 0.000497 17.99 0.38096 0.0 1.870 63.84 3.8E + 2 0.000181 25.62 0.37944 0.0 1.880 1.3E+2 7.7E + 2 3.18E 5 − 61.31 0.37822 0.0 1.888 7.3E+2 4.4E + 3 0.0 62.95 0.37821 0.0 1.888 7.7E+2 4.6E + 3 0.0 64.74 0.37820 0.0 1.888 8.2E+2 4.9E + 3 0.0 66.69 0.37818 0.0 1.888 8.7E+2 5.2E + 3 0.0 68.83 0.37817 0.0 1.888 9.2E+2 5.5E + 3 0.0 71.18 0.37816 0.0 1.889 9.9E+2 5.9E + 3 0.0 73.80 0.37814 0.0 1.889 1.1E+3 6.4E + 3 0.0 76.72 0.37813 0.0 1.889 1.1E+3 6.9E + 3 0.0 80.02 0.37812 0.0 1.889 1.2E+3 7.5E + 3 0.0 83.79 0.37810 0.0 1.889 1.4E+3 8.2E + 3 0.0

Table 5.4: Table for shock propagating from a suddenly opened valve (k=1.3)

0 0 Ty Py P0y Mx My Mx My Tx Px P0x 1.0058 0.99427 0.0 0.010 1.003 1.013 1.00000 1.012 0.98857 0.0 0.020 1.006 1.026 1.00000 1.017 0.98290 0.0 0.030 1.009 1.040 0.99999 1.023 0.97726 0.0 0.040 1.012 1.054 0.99998 1.029 0.97166 0.0 0.050 1.015 1.067 0.99997 120 CHAPTER 5. NORMAL SHOCK Table 5.4: Table for shock propagating from a suddenly opened valve (k=1.3)

0 0 Ty Py P0y Mx My Mx My Tx Px P0x 1.035 0.96610 0.0 0.060 1.018 1.081 0.99995 1.042 0.96056 0.0 0.070 1.021 1.096 0.99991 1.048 0.95506 0.0 0.080 1.024 1.110 0.99987 1.054 0.94959 0.0 0.090 1.028 1.125 0.99981 1.060 0.94415 0.0 0.100 1.031 1.140 0.99975 1.126 0.89159 0.0 0.200 1.063 1.302 0.99792 1.197 0.84227 0.0 0.300 1.098 1.489 0.99288 1.275 0.79611 0.0 0.400 1.136 1.706 0.98290 1.359 0.75301 0.0 0.500 1.177 1.959 0.96631 1.452 0.71284 0.0 0.600 1.223 2.252 0.94156 1.553 0.67546 0.0 0.700 1.274 2.595 0.90734 1.663 0.64073 0.0 0.800 1.333 2.997 0.86274 1.785 0.60847 0.0 0.900 1.400 3.471 0.80734 1.919 0.57853 0.0 1.00 1.478 4.034 0.74136 2.069 0.55074 0.0 1.100 1.570 4.707 0.66575 2.236 0.52495 0.0 1.200 1.681 5.522 0.58223 2.426 0.50100 0.0 1.300 1.815 6.523 0.49333 2.644 0.47875 0.0 1.400 1.980 7.772 0.40226 2.898 0.45807 0.0 1.500 2.191 9.367 0.31281 3.202 0.43882 0.0 1.600 2.467 11.46 0.22904 3.576 0.42089 0.0 1.700 2.842 14.32 0.15495 4.053 0.40418 0.0 1.800 3.381 18.44 0.093988 4.109 0.40257 0.0 1.810 3.448 18.95 0.088718 4.166 0.40097 0.0 1.820 3.519 19.49 0.083607 4.225 0.39938 0.0 1.830 3.592 20.05 0.078654 4.286 0.39780 0.0 1.840 3.669 20.64 0.073863 4.349 0.39624 0.0 1.850 3.749 21.25 0.069233 4.415 0.39468 0.0 1.860 3.834 21.90 0.064766 4.482 0.39314 0.0 1.870 3.923 22.58 0.060462 4.553 0.39160 0.0 1.880 4.016 23.30 0.056322 4.611 0.39037 0.0 1.888 4.096 23.91 0.053088 4.612 0.39035 0.0 1.888 4.097 23.91 0.053053 4.613 0.39034 0.0 1.888 4.098 23.92 0.053018 4.613 0.39033 0.0 1.888 4.099 23.93 0.052984 4.614 0.39031 0.0 1.888 4.099 23.93 0.052949 4.615 0.39030 0.0 1.889 4.100 23.94 0.052914 4.615 0.39029 0.0 1.889 4.101 23.95 0.052879 4.616 0.39027 0.0 1.889 4.102 23.95 0.052844 4.616 0.39026 0.0 1.889 4.103 23.96 0.052809 5.9. TABLES OF NORMAL SHOCKS, K = 1.4 IDEAL GAS 121 Table 5.4: Table for shock propagating from a suddenly opened valve (k=1.3)

0 0 Ty Py P0y Mx My Mx My Tx Px P0x 4.617 0.39025 0.0 1.889 4.104 23.97 0.052775 122 CHAPTER 5. NORMAL SHOCK CHAPTER 6

Normal Shock in Variable Duct Areas

In the previous two chapters, the flow in a variable area duct and a normal shock (discontinuity) were discussed. A discussion of the occurrences of shock in flow in a variable is presented. As it is was presented before, the shock can occur only in steady state when there is a supersonic flow. but also in steady state cases when there is no supersonic flow (in stationary coordinates). As it was shown in Chapter 5, the gas has to pass through a converging–diverging nozzle to obtain a supersonic flow. In the previous chapter, the flow in a convergent– divergent nuzzle was presented when the pressure ratio was above or below the special range. This Chapter will present the flow in this special range of pressure ratios. It is c §©

interesting to note that a normal a shock must occur in these situ- Subsonic#!$¡% ations (pressure ratios). subsonic flow after In Figure (6.1) the re- a shock d Supersonic duced pressure distribution in ! ¡" b the converging–diverging noz- distance, x zle is shown in its whole range of pressure ratios. When the Fig. 6.1: The ﬂow in the nozzle with different back pres- pressure ratio, PB is between sures

123 124 CHAPTER 6. NORMAL SHOCK IN VARIABLE DUCT AREAS point “a” and point “b” the flow is different from what was discussed before. In this case, no continuous pressure possibly can exists. Only in one point where P B = Pb continuous pressure exist. If the back pressure, PB is smaller than Pb a discontinuous point (a shock) will occur. In conclusion, once the flow becomes supersonic, only exact geometry can achieve continuous pressure flow. In the literature, some refer to a nozzle with an area ratio such point b as above the back pressure and it is referred to as an under–expanded nozzle. In the under–expanded case, the nozzle doesn’t provide the maximum thrust possible. On the other hand, when the nozzle exit area is too large a shock will occur and other phenomenon such as plume will separate from the wall inside the nozzle. This nozzle is called an over–expanded nozzle. In comparison of nozzle performance for rocket and aviation, the over–expanded nozzle is worse than the under–expanded nozzle because the nozzle’s large exit area results in extra drag. The location of the shock is determined by geometry to achieve the right back pressure. Obviously if the back pressure, P B, is lower than the critical value (the only value that can achieve continuous pressure) a shock occurs outside of the nozzle. If the back pressure is within the range of Pa to Pb than the exact location determines that after the shock the subsonic branch will match the back pressure.

The ﬁrst example is

Z\[ ]_^8`Aa bcFdOe for academic reasons. It troat XY

has to be recognized that the -£.0/21 shock wave isn’t easily visible &'(*),+ (see Mach’s photography tech- 354687292:2; exit niques). Therefore, this exam- point "e" ple provides a demonstration of

the calculations required for the x y

<¥=>8?A@ BDCFEHG

P§Q*RAS TDUFVHW location even if it isn’t realis- IJHKMLON tic. Nevertheless, this example will provide the fundamentals to Fig. 6.2: A nozzle with normal shock explain the usage of the tools (equations and tables) that were developed so far.

Example 6.1: A large tank with compressed air is attached into a converging–diverging nozzle at 2 pressure 4[Bar] and temperature of 35[◦C]. Nozzle throat area is 3[cm ] and the exit area is 9[cm2]. The shock occurs in a location where the cross section area is 6[cm2]. Calculate the back pressure and the temperature of the ﬂow. (It should be noted that the temperature of the surrounding is irrelevant in this case.) Also determine the critical points for the back pressure (point “a” and point “b”).

SOLUTION Since the key word “large tank” was used that means that the stagnation temperature and pressure are known and equal to the conditions in the tank. 125 First, the exit Mach number has to be determined. This Mach number can be calculated by utilizing the isentropic relationship from the large tank to the shock (point “x”). Then the relationship developed for the shock can be utilized to calculate the Mach number after the shock, (point “y”). From the Mach number after the shock, My, the Mach number at the exit can be calculated by utilizing the isentropic relationship. It has to be realized that for a large tank, the inside conditions are essentially the stagnation conditions (this statement is said without a proof, but can be shown that the correction is negligible for a typical dimension ratio that is over 100. For example, in the case of ratio of 100 the Mach number is 0.00587 and the error is less than %0.1). Thus, the stagnation temperature and pressure are known T0 = 308K and P0 = 4[Bar]. The star area (the throat area), A∗, before the shock is known and given as well. A 6 x = = 2 A∗ 3 With this ratio (A/A∗ = 2) utilizing the Table (5.1) or equation (4.49) or the GDC– Potto, the Mach number, Mx is about 2.197 as shown table below:

M T ρ A P A P ? × T0 ρ0 A P0 A P0 ∗× 2.1972 0.50877 0.18463 2.0000 0.09393 0.18787

With this Mach number, Mx = 2.1972 the Mach number, My can be obtained. From equation (5.22) or from Table (4.2) My ∼= 0.54746. With these values, the subsonic branch can be evaluated for the pressure and temperature ratios.

Ty ρy Py P0y Mx My Tx ρx Px P0x 2.1972 0.54743 1.8544 2.9474 5.4656 0.62941

From Table (4.2) or from equation (4.11) the following Table for the isentropic relationship is obtained

M T ρ A P A P ? × T0 ρ0 A P0 A P0 ∗× 0.54743 0.94345 0.86457 1.2588 0.81568 1.0268

Again utilizing the isentropic relationship the exit conditions can be evaluated. With known Mach number the new star area ratio, Ay/A∗ is known and the exit area can be calculated as A A A 9 e = e y = 1.2588 = 1.8882 A∗ Ay × A∗ × 6 with this area ratio, Ae = 1.8882, one can obtain using the isentropic relationship A∗ as 126 CHAPTER 6. NORMAL SHOCK IN VARIABLE DUCT AREAS

M T ρ A P A P ? × T0 ρ0 A P0 A P0 ∗× 0.32651 0.97912 0.94862 1.8882 0.92882 1.7538

Since the stagnation pressure is constant as well the stagnation temperature, the exit conditions can be calculated.

P P P P P = exit 0 y x P exit P P P P 0 0 y x 0 1 =0.92882 5.466 0.094 4 × 0.81568 × × × ∼=2.34[Bar] The exit temperature is

T T T T T = exit 0 y x T exit T T T T 0 0 y x 0 1 =0.98133 1.854 0.509 308 × 0.951 × × × ∼=299.9K

For the “critical” points ”a” and ”b” are the points that the shock doesn’t occur and yet the ﬂow achieve Mach equal 1 at the throat. In that case we don’t have to go through that shock transition. Yet we have to pay attention that there two possible back pressures that can “achieve” it or target. The area ratio for both cases, is A/A∗ = 3 In the subsonic branch (either using equation or the isentropic Table or GDC-Potto as

M T ρ A P A P ? × T0 ρ0 A P0 A P0 ∗× 0.19745 0.99226 0.98077 3.0000 0.97318 2.9195 2.6374 0.41820 0.11310 3.0000 0.04730 0.14190

Pexit P = P = 0.99226 4 =3.97[Bar] exit P 0 × ∼ 0 For the supersonic sonic branch

Pexit P = P = 0.41820 4 =1.6728[Bar] exit P 0 × ∼ 0 127 It should be noted that the ﬂow rate is constant and maximum for any point beyond the point ”a” even if the shock is exist. The ﬂow rate is expressed as following

P ∗

P ∗  P0 ρ∗ zP0}| { c M=1 P ∗   P0 P ∗   P0 T ∗ m˙ = ρ∗A∗U = A cM =   A √kRT ∗ = A kR T0 RT   T ∗ T0 z }|∗{ R T0 z }| { T0 r z}|{ T ∗ R  T0 T0    T ∗    The temperature and pressure at the| {zthroat} are:

T ∗ T ∗ = T = 0.833 308 = 256.7K T 0 × 0 The temperature at the throat reads

P ∗ P ∗ = P = 0.5283 4 = 2.113[Bar] P 0 × 0 The speed of sound is

c = √1.4 287 256.7 = 321.12[m/sec] × × And the mass ﬂow rate reads

5 410 4 m˙ = 3 10− 321.12 = 0.13[kg/sec] 287 256.7 × × × It is interesting to note that in this case the choking condition is obtained (M = 1) when the back pressure just reduced to less than 5% than original pressure (the pressure in the tank). While the pressure to achieve full supersonic ﬂow through the nozzle the pressure has to be below the 42% the original value. Thus, over 50% of the range of pressure a shock occores some where in the nozzle. In fact in many industrial applications, these kind situations exist. In these applications a small pressure difference can produce a shock wave and a chock ﬂow. For more practical example1 from industrial application point of view.

Example 6.2: In the data from the above example (6.1) where would be shock’s location when the back pressure is 2[Bar]?

1The meaning of the word practical is that in reality the engineer does not given the opportunity to determine the location of the shock but rather information such as pressures and temperature. 128 CHAPTER 6. NORMAL SHOCK IN VARIABLE DUCT AREAS

SOLUTION The solution procedure is similar to what was shown in previous Example (6.1). The solution process starts at the nozzle’s exit and progress to the entrance. The conditions in the tank are again the stagnation conditions. Thus, the exit pressure is between point “a” and point “b”. It follows that there must exist a shock in the nozzle. Mathematically, there are two main possible ways to obtain the solution. In the first method, the previous example information used and expanded. In fact, it requires some iterations by “smart” guessing the different shock locations. The area (location) that the previous example did not “produce” the “right” solution (the exit pressure was 2.113[Bar]. Here, the needed pressure is only 2[Bar] which means that the next guess for the shock location should be with a larger area2. The second (recommended) method is noticing that the flow is adiabatic and the mass flow rate is constant which means that the ratio of the P A∗ = P A∗ 0 × y0 × |@y (upstream conditions are known, see also equation (4.72)).

P A P A 2 9 exit exit = exit exit = × = 1.5[unitless!] P A ∗ P A ∗ 4 3 x0 × x y0 × y × With the knowledge of the ratio P A which was calculated and determines the exit P0A∗ Mach number. Utilizing the Table (4.2) or the GDC-Potto provides the following table is obtained

M T ρ A P A P F ? × T0 ρ0 A P0 A P0 F ∗× ∗ 0.38034 0.97188 0.93118 1.6575 0.90500 1.5000 0.75158

With these values the relationship between the stagnation pressures of the shock are obtainable e.g. the exit Mach number, My, is known. The exit total pressure can be obtained (if needed). More importantly the pressure ratio exit is known. The ratio of the ratio of stagnation pressure obtained by

for Mexit

P0 P0 P 1 2 y = y exit = = 0.5525 P P P 0.905 × 4 0x z }|exit { 0x Looking up in the Table (4.2) or utilizing the GDC-Potto provides

Ty ρy Py P0y Mx My Tx ρx Px P0x 2.3709 0.52628 2.0128 3.1755 6.3914 0.55250

With the information of Mach number (either Mx or My) the area where the shock (location) occurs can be found. First, utilizing the isentropic Table (4.2).

2Of course, the computer can be use to carry this calculations in a sophisticate way. 6.1. NOZZLE EFFICIENCY 129

M T ρ A P A P ? × T0 ρ0 A P0 A P0 ∗× 2.3709 0.47076 0.15205 2.3396 0.07158 0.16747

Approaching the shock location from the upstream (entrance) yields

A 2 A = A∗ = 2.3396 3 = 7.0188[cm ] A × ∼ ∗ Note, as “simple” check this value is larger than the value in the previous example.

6.1 Nozzle efficiency Obviously nozzles are not perfectly efficient and there are several ways to define the nozzleefficiency. One of the effective way is to define the efficiency as the ratio of the energy converted to kinetic energy and the total potential energy could be converted to kinetic energy. The total energy that can be converted is during isentropic process is E = h h (6.1) 0 − exits where hexits is the enthalpy if the flow was isentropic. The actual energy that was used is E = h h (6.2) 0 − exit The efficiency can be defined as

2 h0 hexit (Uactual) η = − = 2 (6.3) h0 hexits (U ) − ideal The typical efficiency of nozzle is ranged between 0.9 to 0.99. In the literature some define also velocity coefficient as the ratio of the actual velocity to the ideal velocity, Vc

2 (Uactual) Vc = √η = 2 (6.4) s (Uideal) There is another less used definition which referred as the coefficient of discharge as the ratio of the actual mass rate to the ideal mass flow rate.

m˙ actual Cd = (6.5) m˙ ideal

6.2 Diffuser Efﬁciency 130 CHAPTER 6. NORMAL SHOCK IN VARIABLE DUCT AREAS P01

The efﬁciency of the diffuser is deﬁned P h 02 as the ratio of the enthalpy change that P2 01 occurred between the entrance to exit 02 stagnation pressure to the kinetic en- 2 P1 ergy.

2(h3 h1) h3 h1 η = −2 = − (6.6) 1 U h0 h1 1 1 − s,entropy For perfect gas equation (6.6) can be converted to Fig. 6.3: Description to clarify the deﬁnition of diffuser efﬁciency 2Cp(T3 T1) η = 2− (6.7) U1 And further expanding equation (6.7) results in

kR T3 2 T3 k 1 2 T1 1 1 −k k 1 T1 k 1 T1 2 T − − − − 3 η = 2 = 2 = 2 1 (6.8) c 2M M M (k 1) T1 − 1 1 1 1 − ! Example 6.3: A wind tunnel combined from a nozzle and a diffuser (actu- nozzle Diffuser ally two nozzles connected by a

1 i£jMk constant area see Figure (6.4)) f£gMh 2 3 4 the required condition at point 3 are: M = 3.0 and pres- capacitor sure of 0.7[Bar] and temperature of 250K. The cross section in area between the nuzzle Compressor and diffuser is 0.02[m2]. What is area of nozzle’s throat and what cooler is area of the diffuser’s throat to maintain chocked diffuser with subsonic ﬂow in the expansion heat section. k = 1.4 can be as- out sumed. Assume that a shock occurs in the test section. Fig. 6.4: Schematic of a supersonic tunnel in a continuous region (and also for example (6.3) SOLUTION The condition at M = 3 is summarized in following table

M T ρ A P A P F ? × T0 ρ0 A P0 A P0 F ∗× ∗ 3.0000 0.35714 0.07623 4.2346 0.02722 0.11528 0.65326 6.2. DIFFUSER EFFICIENCY 131 The nozzle area can be calculated by

? A 2 A∗ = A = 0.02/4.2346 = 0.0047[m ] n A

In this case, P0A∗ is constant (constant mass ﬂow). First the stagnation behind the shock will be

Ty ρy Py P0y Mx My Tx ρx Px P0x 3.0000 0.47519 2.6790 3.8571 10.3333 0.32834

P0n 1 3 A∗d = A∗n 0.0047 0.0143[m ] P0d ∼ 0.32834 ∼

Example 6.4: A shock is moving at 200 [m/sec] in pipe with gas with k = 1.3, pressure of 2[Bar] and temperature of 350K. Calculate the conditions after the shock.

SOLUTION This is a case of completely and suddenly open valve with the shock velocity, temperature and pressure “upstream” known. In this case Potto–GDC provides the following table

0 0 Ty Py P0y Mx My Mx My Tx Px P0x 5.5346 0.37554 0.0 1.989 5.479 34.50 0.021717

The calculations were carried as following: First calculate the Mx as

Mx = U / (k 287. T ) s ∗ ∗ x p Then calculate the My by using Potto-GDC or utilize the Tables. For example Potto-GDC (this code was produce by the program)

Ty ρy Py P0y Mx My Tx ρx Px P0x 5.5346 0.37554 5.4789 6.2963 34.4968 0.02172

The calculation of the temperature and pressure ratio also can be obtain by the same manner. The “downstream” shock number is

Us Msy = 2.09668 Ty ∼ k 287. Tx ∗ ∗ ∗ Tx r 132 CHAPTER 6. NORMAL SHOCK IN VARIABLE DUCT AREAS Finally utilizing the equation to calculate the following

M 0 = M M = 2.09668 0.41087 1.989 y sy − y − ∼

Example 6.5: An inventor interested in a design of tube and piston so that the pressure is doubled in the cylinder when the piston is moving suddenly. The propagating piston is assumed to move into media with temperature of 300K and atmospheric pressure of 1[Bar]. If the steady state is achieved, what will be the piston velocity?

SOLUTION This is an open valve case in which the pressure ratio is given. For this pressure ratio of Py/Px = 2 the following table can be obtained or by using Potto–GDC

Ty ρy Py P0y Mx My Tx ρx Px P0x 1.3628 0.75593 1.2308 1.6250 2.0000 0.96697

The temperature ratio and the Mach numbers for the velocity of the air (and the piston) can be calculated. The temperature at “downstream” (close to the piston) is Ty Ty = Tx = 300 1.2308 = 369.24[◦C] Tx × The velocity of the piston is then U = M c = 0.75593 √1.4 287 369.24 291.16[m/sec] y y ∗ y ∗ ∗ ∗ ∼

Example 6.6: A ﬂow of gas is brought into a sudden stop. The mass ﬂow rate of the gas is 2 [kg/sec] and cross section A = 0.002[m3]. The imaginary gas conditions are temperature is 350K and pressure is 2[Bar] and R = 143[j/kg K] and k = 1.091 (Butane?). Calculate the conditions behind the shock wave.

SOLUTION This is the case of a closed valve in which mass ﬂow rate with the area given. Thus, the “upstream” Mach is given. m˙ mR˙ T 2 287 350 U 0 = = = × × 502.25[m/sec] x ρA P A 200000 0.002 ∼ × 0 Thus the static Mach number, Mx is

0 0 Ux 502.25 Mx = = 2.15 cx √1.091 143 350 ∼ × × 6.2. DIFFUSER EFFICIENCY 133 With this value for the Mach number Potto-GDC provides

0 0 Ty Py P0y Mx My Mx My Tx Px P0x 2.9222 0.47996 2.1500 0.0 2.589 9.796 0.35101

This table was obtained by using the procedure described in this book. The iteration of the procedure are

Ty Py 0 i Mx My My Tx Px 0 3.1500 0.46689 2.8598 11.4096 0.0 1 2.940 0.47886 2.609 9.914 0.0 2 2.923 0.47988 2.590 9.804 0.0 3 2.922 0.47995 2.589 9.796 0.0 4 2.922 0.47996 2.589 9.796 0.0 5 2.922 0.47996 2.589 9.796 0.0 134 CHAPTER 6. NORMAL SHOCK IN VARIABLE DUCT AREAS CHAPTER 7

Nozzle Flow With External Forces

This chapter is under heavy construction. Please ignore. If you want to contribute and add any results of experiments, to this chapter, please do so. You can help especially if you have photos showing these effects.

In the previous chapters a simple model describing the flow in nozzle was explained. In cases where more refined calculations have to carried the gravity or other forces have to be taken into account. Flow in a vertical or horizontal nozzle are different because the gravity. The simplified models that suggests them–self are: friction and adiabatic, isothermal, seem the most applicable. These models can served as limiting cases for more realistic flow.

The effects of the gravity of the nozzle ﬂow in two models isentropic and isothermal is analyzed here. The isothermal nozzle model is suitable in cases where the ﬂow is relatively slow (small Eckert numbers) while as the isentropic model is more suitable for large Eckert numbers.

The two models produces slightly different equations. The equations results in slightly different conditions for the chocking and different chocking speed. Moreover, the working equations are also different and this author isn’t aware of material in the literature which provides any working table for the gravity effect.

135 136 CHAPTER 7. NOZZLE FLOW WITH EXTERNAL FORCES 7.1 Isentropic Nozzle (Q = 0) The energy equation for isentropic nozzle provides

external work or potential difference, i.e. z g × dh + UdU = f(x)dx (7.1) z }| {

7.2 Isothermal Nozzle (T = constant) CHAPTER 8

Isothermal Flow

In this chapter a model dealing with gas that ﬂows through a long tube is described. This model has a applicability to situations which occur in a relatively long distance and where heat transfer is relatively rapid so that the temperature can be treated, for engineering purposes, as a constant . For example, this model is applicable To put discussion for what the when a natural gas ﬂows over several hundreds of meters. Such situations are “relatively rapid” means. common in large cities in U.S.A. where natural gas is used for heating. It is more predominant (more applicable) in situations where the gas is pumped over a length of kilometers.

The high speed of the gas is ob-

tained or explained by the combination vxwzy{v of heat transfer and the friction to the flow l

direction |~}z|

ﬂow. For a long pipe, the pressure dif- z {

m nFoqpsrut

ference reduces the density of the gas. z !!qs

For instance, in a perfect gas, the density is inverse of the pressure (it has c.v. to be kept in mind that the gas undergoes an isothermal process.). To main- Fig. 8.1: Control volume for isothermal ﬂow tain conservation of mass, the velocity increases inversely to the pressure. At critical point the velocity reaches the speed of sound at the exit and hence the ﬂow will be choked1.

1This explanation is not correct as it will be shown later on. Close to the critical point (about, 1/√k, the heat transfer, is relatively high and the isothermal ﬂow model is not valid anymore. Therefore, the study of the isothermal ﬂow above this point is only an academic discussion but also provides the upper limit for Fanno Flow.

137 138 CHAPTER 8. ISOTHERMAL FLOW 8.1 The Control Volume Analysis/Governing equations Figure (8.1) describes the ﬂow of gas from the left to the right. The heat transfer up stream (or down stream) is assumed to be negligible. Hence, the energy equation can be written as the following:

dQ U 2 = c dT + d = c dT (8.1) m˙ p 2 p 0

The momentum equation is written as the following

AdP τ dA = mdU˙ (8.2) − − w wetted area where A is the cross section area (it doesn’t have to be a perfect circle; a close Perhaps more quantitative enough shape is sufﬁcient.). The shear stress is the force per area that acts on the discussions about how “circular” the shape should be. ﬂuid by the tube wall. The Awetted area is the area that shear stress acts on. The second law of thermodynamics reads s s T k 1 P 2 − 1 = ln 2 − ln 2 (8.3) Cp T1 − k P1

The mass conservation is reduced to

m˙ = constant = ρUA (8.4)

Again it is assumed that the gas is a perfect gas and therefore, equation of state is expressed as the following:

P = ρRT (8.5)

8.2 Dimensionless Representation In this section the equations are transformed into the dimensionless form and presented as such. First it must be recalled that the temperature is constant and it seems obvious to write this therefore, equation of state reads equation perhaps to consult with others. dP dρ = (8.6) P ρ

It is convenient to deﬁne a hydraulic diameter

4 Cross Section Area DH = × (8.7) wetted perimeter 8.2. DIMENSIONLESS REPRESENTATION 139 Now, the Fanning friction factor2 is introduced, this factor is a dimensionless friction factor sometimes referred to as the friction coefﬁcient as τ f = w 1 2 (8.8) 2 ρU

Substituting equation (8.8) into momentum equation (8.2) yields

m˙ A 4dx 1 dP f ρU 2 = ρU dU (8.9) − − DH 2 z}|{ Rearranging equation (8.9) and using the identify for perfect gas M 2 = ρU 2/kP yields:

dP 4fdx kP M 2 kP M 2dU = (8.10) − P − D 2 U H Now the pressure, P as a function of the Mach number has to substitute along with velocity, U.

U 2 = kRTM 2 (8.11)

Differentiation of equation (8.11) yields

d(U 2) = kR M 2 dT + T d(M 2) (8.12)

d(M 2) d(U 2) dT = (8.13) M 2 U 2 − T

Now it can be noticed that dT = 0 for isothermal process and therefore

d(M 2) d(U 2) 2U dU 2dU = = = (8.14) M 2 U 2 U 2 U

The dimensionalization of the mass conservation equation yields

dρ dU dρ 2UdU dρ d(U 2) + = + = + = 0 (8.15) ρ U ρ 2U 2 ρ 2 U 2

Differentiation of the isotropic (stagnation) relationship of the pressure (4.11) yields

where are the stagnation equations? put them in a ta- 2 ble It should be noted that Fanning factor based on hydraulic radius, instead of diameter friction equa- put explanation how to derive tion, thus “Fanning f” values are only 1/4th of “Darcy f” values. this expression. 140 CHAPTER 8. ISOTHERMAL FLOW

dP dP 1 kM 2 dM 2 0 = + 2 (8.16) P P k 1 2 M 2 0 1 + −2 M !

Differentiation of equation (4.9) yields: k 1 k 1 dT = dT 1 + − M 2 + T − dM 2 (8.17) 0 2 2 Notice that dT = 0 in an isothermal flow. There is no change in the actual 0 6 temperature of the flow but the stagnation temperature increases or decreases depending on the Mach number (supersonic flow of subsonic flow). Substituting T for equation (8.17) yields:

k 1 2 2 T0 − d M M dT = 2 (8.18) 0 k 1 2 M 2 1 + −2 M Rearranging equation (8.18) yields dT (k 1) M 2 dM 2 0 = − (8.19) T k 1 M 2 0 2 1 + −2 By utilizing the momentum equation it is possible to obtain a relation between the pressure and density. Recalling that an isothermal ﬂow (T = 0) and combining it with perfect gas model yields dP dρ = (8.20) P ρ From the continuity equation (see equation (8.14)) leads dM 2 2dU = (8.21) M 2 U

The four equations momentum, continuity (mass), energy, state are described above. There are 4 unknowns (M, T, P, ρ)3 and with these four equations the solution is attainable. One can notice that there are two possible solutions (because of the square power). These different solutions are supersonic and subsonic solution. 4fL The distance friction, D , is selected as the choice for the independent 4fL variable. Thus, the equations need to be obtained as a function of D . The density is eliminated from equation (8.15) when combined with equation (8.20) to become dP dU = (8.22) P − U 3Assuming the upstream variables are known. 8.2. DIMENSIONLESS REPRESENTATION 141 After substituting the velocity (8.22) into equation (8.10), one can obtain dP 4fdx kP M 2 dP = kP M 2 (8.23) − P − D 2 P H Equation (8.23) can be rearranged into dP dρ dU 1 dM 2 kM 2 dx = = = = 4f (8.24) P ρ − U −2 M 2 −2 (1 kM 2) D − Similarly or by other path the stagnation pressure can be expressed as a function 4fL of D dP kM 2 1 k+1 M 2 dx 0 = − 2 4f (8.25) 2 k 1 2 P0 2 (kM 1) 1 + − M D − 2 dT k (1 k) M 2 dx 0 = − 4f (8.26) 2 k 1 2 T0 2 (1 kM ) 1 + − M D − 2 The variables in equation (8.24) can beseparated to obtain integrable form as follows L 4fdx 1/k 1 kM2 = − dM 2 (8.27) D 2 kM2 Z0 ZM It can be noticed that at the entrance (x = 0) for which M = Mx=0 (the initial 4fL velocity in the tube isn’t zero). The term D is positive for any x, thus, the term on the other side has to be positive as well. To obtain this restriction 1 = kM 2. Thus, the value M = 1 is the limiting case from a mathematical point of view. When √k Mach number larger than M > 1 it makes the right hand side of the integrate √k negative. The physical meaning of this value is similar to M = 1 choked flow which was discussed in a variable area flow in Chapter (4). Further it can be noticed from equation (8.26) that when M 1 the value → √k of right hand side approaches infinity ( ). Since the stagnation temperature (T ) ∞ 0 has a finite value which means that dT . Heat transfer has a limited value 0 → ∞ therefore the model of the flow must be changed. A more appropriate model is an adiabatic flow model yet it can serve as a bounding boundary (or limit). Integration of equation (8.27) yields 1 kM 2 4fLmax = − + ln kM 2 (8.28) D kM 2

The deﬁnition for perfect gas yields M 2 = U 2/kRT and noticing that T = constant is used to describe the relation of the properties at M = 1/√k. By denoting the superscript symbol for the choking condition, one can obtain that ∗ M 2 1/k 2 = 2 (8.29) U U ∗ 142 CHAPTER 8. ISOTHERMAL FLOW Rearranging equation (8.29) is transfered into U = √kM (8.30) U ∗ Utilizing the continuity equation provides ρ 1 ρU = ρ∗U ∗; = = (8.31) ⇒ ρ∗ √kM Reusing the perfect–gas relationship P ρ 1 = = (8.32) P ∗ ρ∗ √kM Now utilizing the relation for stagnated isotropic pressure one can obtain

k k 1 k 1 2 − P0 P 1 + −2 M = k 1 (8.33) P0∗ P ∗ " 1 + 2−k #

Substituting for P equation (8.32) and rearranging yields P ∗

k k k 1 k 1 P0 1 2k − k 1 − 1 = 1 + − M 2 (8.34) P ∗ √k 3k 1 2 M 0 − And the stagnation temperature at the critical point can be expressed as

k 1 2 T0 T 1 + −2 M 2k k 1 2 = k 1 = 1 + − M (8.35) T ∗ T 1 + 3k 1 2 0 ∗ 2−k − These equations (8.30)-(8.35) are presented on in Figure (8.2)

8.3 The Entrance Limitation of Supersonic Branch Situations where the conditions at the tube exit have not arrived at the critical conditions are discussed here. It is very useful to obtain the relationship between the entrance and the exit condition for this case. Denote 1 and 2 as the conditions at the inlet and exit respectably. From equation (8.24)

2 2 2 4fL 4fLmax 4fLmax 1 kM1 1 kM2 M1 D = D D = − 2 − 2 + ln (8.36) 1 − 2 kM1 − kM2 M2

For the case that M1 >> M2 and M1 1 equation (8.36) is reduced into the → following approximation

0 ∼ 2 4fL 1 kM2 D = 2 ln M1 1 − 2 (8.37) − − z kM}|2 { 8.4. COMPARISON WITH INCOMPRESSIBLE FLOW 143 Isothermal Flow

* * * P/P , ρ/ρ and T0/T0 as a function of M

1e+02 4fL  D P ρ * or ∗ P ρ 1e+01 * T0/T0 * P0/P0

0.1

0.01 0.1 1 10 Fri Feb 18 17:23:43 2005 Mach number Fig. 8.2: Description of the pressure, temperature relationships as a function of the Mach number for isothermal ﬂow

Solving for M1 results in

4fL 1 +1 2 D M „ « (8.38) 1 ∼ e This relationship shows the maximum limit that Mach number can approach when 4fL the heat transfer is extraordinarily fast. In reality, even small D > 2 results in a Mach number which is larger than 4.5. This velocity requires a large entrance length to achieve good heat transfer. With this conflicting mechanism obviously the flow is closer to the Fanno flow model. Yet this model provides the directions of the heat transfer effects on the flow.

8.4 Comparison with Incompressible Flow The Mach number of the flow in some instances is relatively small. In these cases, one should expect that the isothermal flow should have similar characteristics as 144 CHAPTER 8. ISOTHERMAL FLOW incompressible flow. For incompressible flow, the pressure loss is expressed as follows

U 2 P P = 4fL (8.39) 1 − 2 D 2

4fL Now note that for incompressible ﬂow U1 = U2 = U and D represent the ratio of the traditional h12. To obtain a similar expression for isothermal ﬂow, a relationship between M2 and M1 and pressures has to be derived. From equation (8.39) one can obtained that

P1 M2 = M1 (8.40) P2

Substituting this expression into (8.40) yields

2 2 4fL 1 P2 P2 D = 2 1 ln (8.41) kM − P1 − P1 1 ! Because f is always positive there is only one solution to the above equation even though M2. Expanding the solution for small pressure ratio drop, P P /P , by some 1 − 2 1 mathematics. denote P P χ = 1 − 2 (8.42) P1

Now equation (8.41) can be transformed into

2 1 P P + P 2 1 4fL = 1 2 − 1 1 ln (8.43) D 2 P2 kM1 − P1 − ! P1 !

2 4fL 1 2 1 D = 2 1 (1 χ) ln (8.44) kM1 − − − 1 χ −

1 1 2 4fL = 2χ χ2 ln (8.45) D kM 2 − − 1 χ 1 − now we have to expand into a series around χ = 0 and remember that

2 x 3 f(x) = f(0) + f 0(0)x + f 00(0) + 0 x (8.46) 2 8.5. SUPERSONIC BRANCH 145 and for example the ﬁrst derivative of

d 1 2 ln = dχ 1 χ − χ=0 2 3 (1 χ) ( 2)(1 χ)− ( 1) = 2 (8.47) − × − − − χ=0

similarly it can be shown that f 00(χ = 0) = 1 equation (8.45) now can be approximated as

4fL 1 2 2 3 D = 2 (2χ χ ) 2χ χ + f χ (8.48) kM1 − − − rearranging equation (8.48) yields

4fL χ 2 3 D = 2 (2 χ) kM1 (2 χ) + f χ (8.49) kM1 − − − and further rearrangement yields

4fL χ 2 2 3 D = 2 2(1 kM1 ) 1 + kM1 χ + f χ (8.50) kM1 − − in cases that χ is small

4fL χ 2 2 D 2 2(1 kM1 ) 1 + kM1 χ (8.51) ≈ kM1 − −

The pressure difference can be plotted as a function of the M1 for given 4fL value of D . Equation (8.51) can be solved explicitly to produce a solution for

2 2 2 1 kM1 1 kM1 kM1 4fL χ = − 2 − 2 2 D (8.52) 1 + kM1 − s1 + kM1 − 1 + kM1

A few observations can be made about equation (8.52).

8.5 Supersonic Branch Apparently, this analysis/model is over simpliﬁed for the supersonic branch and does not produce reasonable results since it neglects to take into account the heat transfer effects. A dimensionless analysis4 demonstrates that all the common materials that the author is familiar which creates a large error in the fundamental

4This dimensional analysis is a bit tricky, and is based on estimates. Currently and ashamedly the author is looking for a more simplified explanation. The current explanation is correct but based on hands waving and definitely does not satisfy the author. 146 CHAPTER 8. ISOTHERMAL FLOW assumption of the model and the model breaks. Nevertheless, this model can provide a better understanding to the trends and deviations of the Fanno flow model. In the supersonic flow, the hydraulic entry length is very large as will be shown below. However, the feeding diverging nozzle somewhat reduces the required entry length (as opposed to converging feeding). The thermal entry length is in the order of the hydrodynamic entry length (look at the Prandtl number, (0.7- 1.0), value for the common gases.). Most of the heat transfer is hampered in the sublayer thus the core assumption of isothermal flow (not enough heat transfer so the temperature isn’t constant) breaks down5. The flow speed at the entrance is very large, over hundred of meters per 4fL second. For example, a gas flows in a tube with D = 10 the required entry Mach number is over 200. Almost all the perfect gas model substances dealt with in this book, the speed of sound is a function of temperature. For this illustration, for most gas cases the speed of sound is about 300[m/sec]. For example, even with low temperature like 200K the speed of sound of air is 283[m/sec]. So, even 4fL for relatively small tubes with D = 10 the inlet speed is over 56 [km/sec]. This requires that the entrance length to be larger than the actual length of the tub for air. Remember from Fluid Dynamic book UD L = 0.06 (8.53) entrance ν The typical values of the the kinetic viscosity, ν, are 0.0000185 kg/m-sec at 300K and 0.0000130034 kg/m-sec at 200K. Combine this information with our case of 4fL D = 10 L entrance = 250746268.7 D On the other hand a typical value of friction coefficient f = 0.005 results in L 10 max = = 500 D 4 0.005 × The fact that the actual tube length is only less than 1% of the entry length means that the assumption is that the isothermal flow also breaks (as in a large response time). Now, if Mach number is changing from 10 to 1 the kinetic energy change is about T0 = 18.37 which means that the maximum amount of energy is insufficient. T0∗ Now with limitation, this topic will be covered in the next version because it provide some insight and boundary to the Fanno Flow model.

8.6 Figures and Tables

5see Kays and Crawford “Convective Heat Transfer” (equation 12-12). 8.7. ISOTHERMAL FLOW EXAMPLES 147 Table 8.1: The Isothermal Flow basic parameters

M 4fL P P0 ρ T0 D P∗ P0∗ ρ∗ T0∗ 0.03000 785.97 28.1718 17.6651 28.1718 0.87516 0.04000 439.33 21.1289 13.2553 21.1289 0.87528 0.05000 279.06 16.9031 10.6109 16.9031 0.87544 0.06000 192.12 14.0859 8.8493 14.0859 0.87563 0.07000 139.79 12.0736 7.5920 12.0736 0.87586 0.08000 105.89 10.5644 6.6500 10.5644 0.87612 0.09000 82.7040 9.3906 5.9181 9.3906 0.87642 0.10000 66.1599 8.4515 5.3334 8.4515 0.87675 0.20000 13.9747 4.2258 2.7230 4.2258 0.88200 0.25000 7.9925 3.3806 2.2126 3.3806 0.88594 0.30000 4.8650 2.8172 1.8791 2.8172 0.89075 0.35000 3.0677 2.4147 1.6470 2.4147 0.89644 0.40000 1.9682 2.1129 1.4784 2.1129 0.90300 0.45000 1.2668 1.8781 1.3524 1.8781 0.91044 0.50000 0.80732 1.6903 1.2565 1.6903 0.91875 0.55000 0.50207 1.5366 1.1827 1.5366 0.92794 0.60000 0.29895 1.4086 1.1259 1.4086 0.93800 0.65000 0.16552 1.3002 1.0823 1.3002 0.94894 0.70000 0.08085 1.2074 1.0495 1.2074 0.96075 0.75000 0.03095 1.1269 1.0255 1.1269 0.97344 0.80000 0.00626 1.056 1.009 1.056 0.98700 0.81000 0.00371 1.043 1.007 1.043 0.98982 0.81879 0.00205 1.032 1.005 1.032 0.99232 0.82758 0.000896 1.021 1.003 1.021 0.99485 0.83637 0.000220 1.011 1.001 1.011 0.99741 0.84515 0.0 1.000 1.000 1.000 1.000

8.7 Isothermal Flow Examples There can be several kinds of questions aside from the proof questions6 Generally, the “engineering” or practical questions can be divided into driving force (pressure difference), resistance (diameter, friction factor, friction coefficient, etc.), and mass flow rate questions. In this model no questions about shock (should) exist7. The driving force questions deal with what should be the pressure difference to obtain certain flow rate. Here is an example.

6The proof questions are questions that ask for proof or for finding a mathematical identity (normally good for mathematicians and study of perturbation methods). These questions or examples will appear in the later versions. 7Those who are mathematically inclined can include these kinds of questions but there are no real world applications to isothermal model with shock. 148 CHAPTER 8. ISOTHERMAL FLOW Example 8.1: A tube of 0.25 [m] diameter and 5000 [m] in length is attached to a pump. What should be the pump pressure so that a flow rate of 2 [kg/sec] will be achieved? Assume that friction factor f = 0.005 and the exit pressure is 1[bar]. The specific J heat for the gas, k = 1.31, surroundings temperature 27◦C, R = 290 Kkg . Hint: calculate the maximum flow rate and then check if this request is reasonabh ile.

SOLUTION If the flow was incompressible then for known density, ρ, the velocity can be calcu- 4fL U 2 lated by utilizing ∆P = D 2g . In incompressible flow, the density is a function of the entrance Mach number. The exit Mach number is not necessarily 1/√k i.e. the flow is not choked. First, check whether flow is choked (or even possible). 4fL Calculating the resistance, D 4 0.0055000 4fL = × = 400 D 0.25 Utilizing Table (8.1) or the program provides

M 4fL P P0 ρ T0 D P∗ P0∗ ρ∗ T0∗ 0.04331 400.00 20.1743 12.5921 0.0 0.89446

The maximum flow rate (the limiting case) can be calculated by utilizing the above table. The velocity of the gas at the entrance U = cM = 0.04331 × √1.31 290 300 = 14.62 m . The density reads × × ∼ sec P 2, 017, 450 kg ρ = = = 23.19 RT 290 300 ∼ m3 × The maximum flow rate then reads π (0.25)2 kg m˙ = ρAU = 23.19 × 14.62 = 16.9 × 4 × ∼ sec The maximum flow rate is larger then the requested mass rate hence the flow is not choked. It is note worthy to mention that since the isothermal model breaks around the choking point, the flow rate is really some what different. It is more appropriate to assume an isothermal model hence our model is appropriate. To solve this problem the flow rate has to be calculated as

kg m˙ = ρAU = 2.0 sec

P1 kU P1 kU P1 m˙ = A = A = AkM1 RT k √kRT √kRT c 8.7. ISOTHERMAL FLOW EXAMPLES 149 Now combining with equation (8.40) yields M P Ak m˙ = 2 2 c mc˙ 2 337.59 M2 = = ×π (0.25)2 = 0.103 P2Ak 100000 × 1.31 × 4 × From Table (8.1) or by utilizing the program

M 4fL P P0 ρ T0 D P∗ P0∗ ρ∗ T0∗ 0.10300 66.6779 8.4826 5.3249 0.0 0.89567

The entrance Mach number is obtained by

4fL = 66.6779 + 400 = 466.68 D 1 ∼

M 4fL P P0 ρ T0 D P∗ P0∗ ρ∗ T0∗ 0.04014 466.68 21.7678 13.5844 0.0 0.89442

The pressure should be P = 21.76780 8.4826 = 2.566[bar] × Note that tables in this example are for k = 1.31

Example 8.2: A flow of gas was considered for a distance of 0.5 [km] (500 [m]). A flow rate of 0.2 [kg/sec] is required. Due to safety concerns, the maximum pressure allowed for the gas is only 10[bar]. Assume that the flow is isothermal and k=1.4, calculate the required diameter of tube. The friction coefficient for the tube can be assumed as 0.02 (A relative smooth tube of cast iron.). Note that tubes are provided in increments of 0.5 [in]8. You can assume that the soundings temperature to be 27◦C.

SOLUTION At ﬁrst, the minimum diameter will be obtained when the ﬂow is choked. Thus, the maximum M1 that can be obtained when the M2 is at its maximum and back pressure is at the atmospheric pressure.

Mmax

P2 1 1 M1 = M2 = = 0.0845 P1 z}|{√k 10 Now, with the value of M1 either by utilizing Table (8.1) or using the provided program yields

8It is unfortunate, but it seems that this standard will be around in USA for some time. 150 CHAPTER 8. ISOTHERMAL FLOW

M 4fL P P0 ρ T0 D P∗ P0∗ ρ∗ T0∗ 0.08450 94.4310 10.0018 6.2991 0.0 0.87625

4fLmax With D = 94.431 the value of minimum diameter. 4fL 4 0.02 500 D = × × 0.42359[m] = 16.68[in] 4fLmax ' 94.43 ' D However, the pipes are provided only in 0.5 increments and the next size is 17[in] or 0.4318[m]. With this pipe size the calculations are to be repeated in reverse and produces: (Clearly the maximum mass is determined with)

P P AM√k m˙ = ρAU = ρAMc = AM√kRT = RT √RT The usage of the above equation clearly applied to the whole pipe. The only point that must be emphasized is that all properties (like Mach number, pressure and 4fL etc) have to be taken at the same point. The new D is 4 0.02 500 4fL = × × 92.64 D 0.4318 '

M 4fL P P0 ρ T0 D P∗ P0∗ ρ∗ T0∗ 0.08527 92.6400 9.9110 6.2424 0.0 0.87627

To check whether the ﬂow rate satisﬁes the requirement

6 π 0.43182 10 × 0.0853 √1.4 m˙ = × 4 × × 50.3[kg/sec] √287 300 ≈ × Since 50.3 0.2 the mass ﬂow rate requirement is satisﬁed. ≥ It should be noted that P should be replaced by P0 in the calculations. The speed of sound at the entrance is m c = √kRT = √1.4 287 300 = 347.2 × × ∼ sec h i and the density is P 1, 000, 000 kg ρ = = = 11.61 RT 287 300 m3 × The velocity at the entrance should be m U = M c = 0.08528 347.2 = 29.6 ∗ × ∼ sec h i 8.7. ISOTHERMAL FLOW EXAMPLES 151 The diameter should be

4m˙ 4 0.2 D = = × = 0.027 πUρ π 29.6 11.61 ∼ s r × × Nevertheless, for the sake of the exercise the other parameters will be calculated. This situation is reversed question. The ﬂow rate is given with the diameter of the pipe. It should be noted that the ﬂow isn’t choked.

Example 8.3: A gas flows of from a station (a) with pressure of 20[bar] through a pipe with 0.4[m] diameter and 4000 [m] length to a different station (b). The pressure at the exit (station (b)) is 2[bar]. The gas and the sounding temperature can be assumed to be 300 K. Assume that the flow is isothermal, k=1.4, and the average friction f=0.01. Calculate the Mach number at the entrance to pipe and the flow rate.

SOLUTION First, the information whether the ﬂow is choked needs to be found. Therefore, at ﬁrst it will be assumed that the whole length is the maximum length.

4 0.01 4000 4fLmax = × × = 400 D 0.4

4fLmax with D = 400 the following can be written

4fL T0 ρ P P0 M T T T T D T0∗ ρ∗ P∗ P0∗ 0.0419 400.72021 0.87531 20.19235 20.19235 12.66915

P0 From the table M1 0.0419 ,and T 12.67 ≈ P0∗ ≈

T 28 P ∗ = 2.21[bar] 0 ∼ 12.67 ' The pressure at point (b) by utilizing the isentropic relationship (M = 1) pressure ratio is 0.52828.

T P0∗ P2 = = 2.21 0.52828 = 1.17[bar] P2 × T P0∗ As the pressure at point (b) is smaller than the actual pressure P ∗ < P2 than the actual pressure one must conclude that the ﬂow is not choked. The solution is an iterative process.

4fL 1. guess reasonable value of M1 and calculate D 152 CHAPTER 8. ISOTHERMAL FLOW

2. Calculate the value of 4fL by subtracting 4fL 4fL D 2 D 1 − D

3. Obtain M2 from the Table ? or by using the Potto–GDC .

4. Calculate the pressure, P2 bear in mind that this isn’t the real pressure but based on the assumption

5. Compare the results of guessed pressure P2 with the actual pressure and choose new Mach number M1 accordingly.

Now the process has been done for you and is provided in Figure (??) or in the table obtained from the provided program.

4fLmax 4fL P2 M1 M2 D 1 D P1 0.0419 0.59338 400.32131 400.00000 0.10000

The ﬂow rate is

P √k π D2 2000000√1.4 m˙ = ρAMc = × M = π 0.22 0.0419 √RT 4 √300 287 × × × 42.46[kg/sec] '

In this chapter, there are no examples on isothermal with supersonic ﬂow.

8.8 Unchoked situation

Table 8.4: The ﬂow parameters for unchoked ﬂow

M M 4fLmax 4fL P2 1 2 P D 1 D 1 0.7272 0.84095 0.05005 0.05000 0.10000

0.6934 0.83997 0.08978 0.08971 0.10000 0.6684 0.84018 0.12949 0.12942 0.10000 0.6483 0.83920 0.16922 0.16912 0.10000 0.5914 0.83889 0.32807 0.32795 0.10000 0.5807 0.83827 0.36780 0.36766 0.10000 0.5708 0.83740 0.40754 0.40737 0.10000 8.8. UNCHOKED SITUATION 153

M1 isothermal flow

0.9

0.8 P2 / P1 = 0.8 0.7 P2 / P1 = 0.5 P2 / P1 = 0.2 0.6 P2 / P1 = 0.10 1

M 0.5

0.4

0.3

0.2

0.1

0 0 10 20 30 40 50 60 70 80 90 100 4fL  Fri Feb 25 17:20:14 2005 D

Fig. 8.3: The Mach number at the entrance to a tube under isothermal ﬂow model as a 4fL function D 154 CHAPTER 8. ISOTHERMAL FLOW CHAPTER 9

Fanno Flow °,±

An adiabatic ﬂow with friction is ¤ named after Ginno Fanno a Jewish 8 flow

engineer. This model is the second direction ¢¡£u

¥¦¨§¥ !D

pipe ﬂow model described here.

¦¨§

is that heat transfer is negligible and ²,³ 1 can be ignored . This model is ap- c.v. plicable to flow processes which are No heat transer very fast compared to heat transfer mechanisms with small Eckert Fig. 9.1: Control volume of the gas flow in a con- number. stant cross section This model explains many industrial flow processes which includes emp- tying of pressured container through a relatively short tube, exhaust system of an internal combustion engine, compressed air systems, etc. As this model raised from need to explain the steam flow in turbines.

9.1 Introduction Consider a gas ﬂowing through a conduit with a friction (see Figure (9.1)). It is advantages to examine the simplest situation and yet without losing the core properties of the process. Later, more general cases will be examined2.

1Even the friction does not convert into heat 2Not ready yet, discussed on the ideal gas model and the entry length issues.

155 156 CHAPTER 9. FANNO FLOW 9.2 Model The mass (continuity equation) balance can be written as

m˙ = ρAU = constant (9.1) , ρ U = ρ U → 1 1 2 2

The energy conservation (under the assumption that this model is adiabatic ﬂow and the friction is not transformed into thermal energy) reads

T01 = T02 (9.2) 2 2 U1 U2 , T1 + = T2 + → 2cp 2cp (9.3)

Or in a derivative form U 2 C dT + d = 0 (9.4) p 2

Again for simplicity, the perfect gas model is assumed3.

P = ρRT (9.5) P P , 1 = 2 → ρ1T1 ρ2T2

It is assumed that the ﬂow can be approximated as one–dimensional. The force acting on the gas is the friction at the wall and the momentum conservation reads

AdP τ dA = mdU˙ (9.6) − − w w

It is convenient to deﬁne a hydraulic diameter as

4 Cross Section Area DH = × (9.7) wetted perimeter

Or in other words πD 2 A = H (9.8) 4

3The equation of state is written again here so that all the relevant equations can be found when this chapter is printed separately. 9.3. NON–DIMENSIONALIZATION OF THE EQUATIONS 157

It is convenient to substitute D for DH and yet it still will be referred to the same name as the hydraulic diameter. The inﬁnitesimal area that shear stress is acting on is

dAw = πDdx (9.9)

Introducing the Fanning friction factor as a dimensionless friction factor which is some times referred to as the friction coefﬁcient and reads as the following: τ f = w 1 2 (9.10) 2 ρU

By utilizing equation (9.2) and substituting equation (9.10) into momentum equation (9.6) yields

A τw m˙ A πD2 1 dP πDdx f ρU 2 = A ρU dU (9.11) − z }|4 { − z 2}| { z}|{ Dividing equation (9.11) by the cross section area, A and rearranging yields

4fdx 1 dP + ρU 2 = ρUdU (9.12) − D 2 The second law is the last equation to be utilized to determine the ﬂow direction.

s s (9.13) 2 ≥ 1

9.3 Non–dimensionalization of the equations Before solving the above equation a dimensionless process is applied. By utilizing the deﬁnition of the sound speed to produce the following identities for perfect gas

U 2 U 2 M 2 = = (9.14) c k RT P ρ |{z} Utilizing the deﬁnition of the perfect gas results in

ρU 2 M 2 = (9.15) kP 158 CHAPTER 9. FANNO FLOW Using the identity in equation (9.14) and substituting it into equation (9.11) and after some rearrangement yields

ρU 2 4fdx 1 ρU 2 dU dP + kP M 2 = dU = kP M 2 (9.16) − DH 2 U U z }| { By further rearranging equation (9.16) results in

dP 4fdx kM 2 dU = kM 2 (9.17) − P − D 2 U It is convenient to relate expressions of (dP/P ) and dU/U in terms of the Mach number and substituting it into equation (9.17). Derivative of mass conservation ((9.2)) results in

dU U dρ 1 dU 2 + = 0 (9.18) ρ z2 }|U 2{ The derivation of the equation of state (9.5) and dividing the results by equation of state (9.5) results dP dρ dT = + (9.19) P ρ dT

Derivation of the Mach identity equation (9.14) and dividing by equation (9.14) yields

d(M 2) d(U 2) dT = (9.20) M 2 U 2 − T

Dividing the energy equation (9.4) by Cp and by utilizing the deﬁnition Mach number yields

dT 1 1 U 2 U 2 + d = T kR T U 2 2 (k 1) − Cp dT (k 1) U 2 U 2 , | +{z −} d = → T kRT U 2 2 c2 dT k 1 dU 2 , +| {z }− M 2 = 0 (9.21) → T 2 U 2

This equation is obtained by Equations (9.17), (9.18), (9.19), (9.20), and (9.21) need to be solved. These equa- combining the deﬁnition of Mach number with equation of state and mass conservation. Thus, the original limitations must be applied to the resulting equation. 9.3. NON–DIMENSIONALIZATION OF THE EQUATIONS 159 tions are separable so one variable is a function of only single variable (the chosen as the independent variable). Explicit explanation is provided for only two variables, 4fL the rest variables can be done in a similar fashion. The dimensionless friction, D , is chosen as the independent variable since the change in the dimensionless re- 4fL sistance, D , causes the change in the other variables. Combining equations (9.19) and (9.21) when eliminating dT/T results dP dρ (k 1)M 2 dU 2 = − (9.22) P ρ − 2 U 2

dρ The term ρ can be eliminated by utilizing equation (9.18) and substituting it into equation (9.22) and rearrangement yields dP 1 + (k 1)M 2 dU 2 = − (9.23) P − 2 U 2 The term dU 2/U 2 can be eliminated by using (9.23) dP kM 2 1 + (k 1)M 2 4fdx = − (9.24) P − 2(1 M 2) D − The second equation for Mach number, M variable is obtained by combining equation (9.20) and (9.21) by eliminating dT/T . Then dρ/ρ and U are eliminated by utilizing equation (9.18) and equation (9.22). The only variable that is left is P (or dP/P ) which can be eliminated by utilizing equation (9.24) and results in 4fdx 1 M 2 dM 2 = − (9.25) D 4 k 1 2 kM (1 + −2 M ) Rearranging equation (9.25) results in

2 2 k 1 2 dM kM 1 + − M 4fdx = 2 (9.26) M 2 1 M 2 D − After similar mathematical manipulation one can get the relationship for the velocity to read dU kM 2 4fdx = (9.27) U 2 (1 M 2) D − and the relationship for the temperature is dT 1 dc k(k 1)M 4 4fdx = = − (9.28) T 2 c − 2(1 M 2) D − density is obtained by utilizing equations (9.27) and (9.18) to obtain dρ kM 2 4fdx = (9.29) ρ −2 (1 M 2) D − 160 CHAPTER 9. FANNO FLOW The stagnation pressure is similarly obtained as

dP kM 2 4fdx 0 = (9.30) P0 − 2 D

The second law reads dT dP ds = C ln R ln (9.31) p T − P

The stagnation temperature expresses as T = T (1 + (1 k)/2M 2). Taking 0 − derivative of this expression when M remains constant yields dT = dT (1 + (1 0 − k)/2M 2) and thus when these equations are divided they yield

dT/T = dT0/T0 (9.32)

In similar fashion the relationship between the stagnation pressure and the pressure can be substituted into the entropy equation and result in

dT0 dP0 ds = Cp ln R ln (9.33) T0 − P0

The ﬁrst law requires that the stagnation temperature remains constant, (dT0 = 0). Therefore the entropy change is

ds (k 1) dP = − 0 (9.34) Cp − k P0

Using the equation for stagnation pressure the entropy equation yields

ds (k 1)M 2 4fdx = − (9.35) Cp 2 D

9.4 The Mechanics and Why the Flow is Choked? The trends of the properties can be examined by looking in equations (9.24) through (9.34). For example, from equation (9.24) it can be observed that the critical point is when M = 1. When M < 1 the pressure decreases downstream as can be seen from equation (9.24) because fdx and M are positive. For the same reasons, in the supersonic branch, M > 1, the pressure increases downstream. This pressure increase is what makes compressible flow so different from “conven- tional” flow. Thus the discussion will be divided into two cases: One, flow above speed of sound. Two, flow with speed below the speed of sound. 9.5. THE WORKING EQUATIONS 161

Why the ﬂow is choked?

Here, the explanation is based on the equations developed earlier and there is no known explanation that is based on the physics. First, it has to be recognized that the critical point is when M = 1 it will show a change in the trend and it is singular point by itself. For example, dP (@M = 1) = and mathematically it is a ∞ singular point (see equation (9.24)). Observing from equation (9.24) that increase or decrease from subsonic just below one M = (1 ) to above just above one − M = (1 + ) requires a change in a sign pressure direction. However, the pressure has to be a monotonic function which means that flow cannot crosses over the point of M = 1. This constrain means that because the flow cannot “crossover” M = 1 the gas has to reach to this speed, M = 1 at the last point. This situation is called choked flow.

The Trends

The trends or whether the variables are increasing or decreasing can be observed from looking at the equation developed. For example, the pressure can be examined by looking at equation (9.26). It demonstrates that the Mach number increases downstream when the ﬂow is subsonic. On the other hand, when the ﬂow is supersonic, the pressure decreases. The summary of the properties changes on the sides of the branch

Subsonic Supersonic Pressure, P decrease increase Mach number, M increase decrease Velocity, U increase decrease Temperature, T decrease increase Density, ρ decrease increase Stagnation Temperature, T0 decrease increase

9.5 The working equations Integration of equation (9.25) yields

4 Lmax 1 1 M 2 k + 1 k+1 M 2 fdx = − + ln 2 (9.36) D k M 2 2k 1 + k 1 M 2 ZL −2

A representative friction factor is deﬁned as

1 Lmax f¯ = fdx (9.37) L max Z0 162 CHAPTER 9. FANNO FLOW By utilizing the mean average theorem equation (9.36) yields

4f¯L 1 1 M 2 k + 1 k+1 M 2 max = − + ln 2 (9.38) D k M 2 2k k 1 2 1 + −2 M

It is common to replace the f¯ with f which is adopted in this book. Equations (9.24), (9.27), (9.28), (9.29), (9.29), and (9.30) can be solved. 4fL For example, the pressure as written in equation (9.23) is represented by D , 4fL and Mach number. Now equation (9.24) can eliminate term D and describe the pressure on the Mach number. Dividing equation (9.24) in equation (9.26) yields

dP 1 + (k 1M 2 P = − dM 2 (9.39) dM 2 − 2 k 1 2 M 2 2M 1 + −2 M The symbol “*” denotes the state when the ﬂow ischoked and Mach number is equal to 1. Thus, M = 1 when P = P ∗ Equation (9.39) can be integrated to yield:

P 1 k+1 = 2 (9.40) P M k 1 2 ∗ s1 + −2 M In the same fashion the variables ratio can be obtained T c2 k+1 = = 2 (9.41) T 2 k 1 2 ∗ c∗ 1 + −2 M

k 1 2 ρ 1 1 + − M = 2 (9.42) ρ M k+1 ∗ s 2

1 k+1 U ρ − = = M 2 (9.43) U ρ 1 + k 1 M 2 ∗ ∗ s −2 The stagnation pressure decreases and can be expressed by

k 1 k 2 k 1 (1+ −2 M ) −

P0 P P 0 = z}|{P (9.44) P0∗ P0∗ P ∗ P ∗ k 2 k 1 ( k+1 ) − |{z} 9.5. THE WORKING EQUATIONS 163 Using the pressure ratio in equation (9.40) and substituting it into equation (9.44) yields

k k 1 k 1 2 − k 1 2 P0 1 + −2 M 1 1 + −2 M = k+1 k+1 (9.45) P ∗ M 0 2 ! s 2 And further rearranging equation (9.45) provides

k+1 2(k 1) k 1 2 − P0 1 1 + −2 M = k+1 (9.46) P ∗ M 0 2 ! The integration of equation (9.34) yields

k+1 k s s∗ k + 1 − = ln M 2 (9.47) v 2 k 1 2 cp u 2M 1 + − M ! u 2 t The results of these equations are plottedin Figure (9.2) The Fanno ﬂow is in discussion about Reynolds number and dimensionless 4fL friction parameter. Fanno Flow D * * * P/P , ρ/ρ and T/T as a function of M

1e+02 4fL  D P * P 1e+01 * T0/T0 * P0/P0 U/U* 1

0.1

0.01 0.1 1 10 Fri Sep 24 13:42:37 2004 Mach number

Fig. 9.2: Various parameters in Fanno ﬂow as a function of Mach number many cases shockless and therefore a relationship between two points should be 164 CHAPTER 9. FANNO FLOW derived. In most times, the “star” values are imaginary values that represent the value at choking. The real ratio can be obtained by two star ratios as an example

T T2 T ∗ M2 = T (9.48) T1 T ∗ M1

A special interest is the equation for the dimensionless friction as following

L2 4fL Lmax 4fL Lmax 4fL dx = dx dx (9.49) D D − D ZL1 ZL1 ZL2 Hence,

4fL 4fL 4fL max = max (9.50) D D − D 2 1

9.6 Examples of Fanno Flow

Example 9.1: à

Air ﬂows from a reservoir and enters a uni- Û£ÜAÝ·Þß

´µ·¶¸ ¶s¹©º »§¼

form pipe with a diameter of 0.05 [m] and ½0¾,¿5À áãâåäÑæ©ç è§é Ç length of 10 [m]. The air exits to the at- ÁÃÂ,Ä_ÅMÆ

mosphere. The following conditions pre-

ÔÕAÖ·×sØsÙ Ú

vail at the exit: P2 = 1[bar] temperature ÏÑÐDÒsÓ 4 ÈÊÉAËÍÌDÎ T2 = 27◦C M2 = 0.9 . Assume that the average friction factor to be f = 0.004 and that the ﬂow from the reservoir up to the Fig. 9.3: Schematic of Example (9.1) pipe inlet is essentially isentropic. Esti- mate the total temperature and total pressure in the reservoir under the Fanno ﬂow model.

SOLUTION For isentropic, the ﬂow to the pipe inlet, the temperature and the total pressure at the pipe inlet are the same as those in the reservoir. Thus, ﬁnding the total pressure and temperature at the pipe inlet is the solution. With the Mach number and temperature known at the exit, the total temperature at the entrance can be 4fL obtained by knowing the D . For given Mach number (M = 0.9) the following is obtained.

M 4fL P P0 ρ U T D P∗ P0∗ ρ∗ U∗ T∗ 0.90000 0.01451 1.1291 1.0089 1.0934 0.9146 1.0327

4This property is given only for academic purposes. There is no Mach meter. 9.6. EXAMPLES OF FANNO FLOW 165 So, the total temperature at the exit is

T ∗ 300 T ∗ = T = = 290.5[K] |2 T 2 1.0327 2

4fL To ”move” to the other side of the tube the D is added as

4fL 4fL 4fL 4 0.004 10 D = D + D = × × + 0.01451 3.21 1 2 0.05 '

4fL The rest of the par ameters can be obtained with the new D either from Table (9.1) by interpolations or by utilizing the attached program.

M 4fL P P0 ρ U T D P∗ P0∗ ρ∗ U∗ T∗ 0.35886 3.2100 3.0140 1.7405 2.5764 0.38814 1.1699

Note that the subsonic branch is chosen. The stagnation ratios has to be added for M = 0.35886

M T ρ A P A P F ? × T0 ρ0 A P0 A P0 F ∗× ∗ 0.35886 0.97489 0.93840 1.7405 0.91484 1.5922 0.78305

The total pressure P01 can be found from the combination of the ratios as follows:

P ∗

z P ∗}| P { P0 P = P 01 2 P P P z }| {2 ∗ 1 1 1 1 =1 3.014 = 2.91[Bar] × 1.12913 × × 0.915

T ∗

z T ∗}| T { T0 T = T 01 2 T T T z }| {2 ∗ 1 1 1 1 =300 1.17 348K = 75◦C × 1 .0327 × × 0.975 '

Another academic question: 166 CHAPTER 9. FANNO FLOW Example 9.2:

A system is composed of a convergent-

êãë§ìMí ìHîOïMð ñ_ò

ó©ô2õö ÷Mø ùHúMû ü ýMþ ÿ

"!"#

divergent nozzle followed by a tube with ¦ ¦

length of 2.5 [cm] in diameter and 1.0 [m] ¢¡¤£¦¥¨§©§¨ shock atmosphere long. The system is supplied by a vessel. conditions The vessel conditions are at 29.65 [Bar], d-c nozzle 400 K. With these conditions a pipe inlet Mach number is 3.0. A normal shock wave Fig. 9.4: The schematic of Example (9.2) occurs in the tube and the ﬂow discharges to the atmosphere, determine: (a) the mass ﬂow rate through the system; (b) the temperature at the pipe exit; and

SOLUTION

(a) Assuming that the pressure vessel is very much larger than the pipe, therefore the velocity in the vessel can be assumed to be small enough so it can be neglected. Thus, the stagnation conditions can be approximated for the condition in the tank. It is further assumed that the ﬂow through the nozzle can be approximated as isentropic. Hence, T01 = 400K and P01 = 29.65[P ar] The mass ﬂow rate through the system is constant and for simplicity point 1 is chosen in which,

m˙ = ρAMc

The density and speed of sound are unknowns and need to be computed. With the isentropic relationship the Mach number at point one (1) is known, then the following can be found either from Table (9.1) or the Potto–GDC

M T ρ A P A P F ? × T0 ρ0 A P0 A P0 F ∗× ∗ 3.0000 0.35714 0.07623 4.2346 0.02722 0.11528 0.65326

The temperature is

T1 T1 = T01 = 0.357 400 = 142.8K T01 × 9.6. EXAMPLES OF FANNO FLOW 167 Using the temperature, the speed of sound can be calculated as

c = √kRT = √1.4 287 142.8 239.54[m/sec] 1 × × ' The pressure at point 1 can be calculated as

P1 P1 = P01 = 0.027 30 0.81[Bar] P01 × ' The density as a function of other properties at point 1 is

P 8.1 104 kg ρ = = × 1.97 1 RT 287 142.8 ' m3 1 ×

The mass ﬂow rate can be evaluated from equation (9.2) π 0.0252 kg m˙ = 1.97 × 3 239.54 = 0.69 × 4 × × sec

(b) First, check whether the ﬂow is shockless by comparing the ﬂow resistance and the maximum possible resistance. From the Table (9.1) or by using the Potto–GDC, to obtain the following

M 4fL P P0 ρ U T D P∗ P0∗ ρ∗ U∗ T∗ 3.0000 0.52216 0.21822 4.2346 0.50918 1.9640 0.42857

and the conditions of the tube are 4 0.005 1.0 4fL = × × = 0.8 D 0.025

Since 0.8 > 0.52216 the ﬂow is choked and with a shock wave. The exit pressure determines the location of the shock, if a shock exists, by comparing “possible” Pexit to PB. Two possibilities are needed to be checked; one, the shock at the entrance of the tube, and two, shock at the exit and comparing the pressure ratios. First, the possibility that the shock wave occurs immediately at the entrance for which the ratio for Mx are (shock wave Table (5.1))

Ty ρy Py P0y Mx My Tx ρx Px P0x 3.0000 0.47519 2.6790 3.8571 10.3333 0.32834

After shock wave the ﬂow is subsonic with “M1”= 0.47519. (Fanno ﬂow Table (9.1)) 168 CHAPTER 9. FANNO FLOW

M 4fL P P0 ρ U T D P∗ P0∗ ρ∗ U∗ T∗ 0.47519 1.2919 2.2549 1.3904 1.9640 0.50917 1.1481

The stagnation values for M = 0.47519 are

M T ρ A P A P F ? × T0 ρ0 A P0 A P0 F ∗× ∗ 0.47519 0.95679 0.89545 1.3904 0.85676 1.1912 0.65326

The ratio of exit pressure to the chamber total pressure is

1 1

P P P P P0 P 2 = 2 ∗ 1 y 0x P P P P P P 0 z }|∗ { 1 0y 0x z }|0 { 1 = 1 0.8568 0.32834 1 × 2.2549 × × × = 0.12476

The actual pressure ratio 1/29.65 = 0.0338 is smaller than the case in which shock occurs at the entrance. Thus, the shock is somewhere downstream. One possible way to find the exit temperature, T2 is by finding the location of the shock. To find the location of the shock ratio of the pressure ratio, P2 is needed. With the location of shock, “claiming” upstream from the P1 exit through shock to the entrance. For example, calculate the parameters for shock location 4fL with known D in the “y” side. Then either by utilizing shock table or the program, to obtain the upstream Mach number. The procedure for the calculations: 1) Calculate the entrance Mach number assuming the shock occurs at the exit:

0 a) set M2 = 1 assume the ﬂow in the entire tube is supersonic: 0 b) calculated M1 Note this Mach number is the high Value. 2) Calculate the entrance Mach assuming shock at the entrance.

a) set M2 = 1 4fL b) add D and calculated M1’ for subsonic branch c) calculated Mx for M1’ Note this Mach number is the low Value. 5 To check Secant Method. 3) According your root ﬁnding algorithm calculate or guess the shock location and then compute as above the new M1.

5You can use any method you which, but be-careful second order methods like Newton-Rapson method can be unstable. 9.7. SUPERSONIC BRANCH 169

a) set M2 = 1 4fL b) for the new D and compute the new My’ for the subsonic branch c) calculated Mx’ for the My’ 4fL d) Add the leftover of D and calculated the M1 4) guess new location for the shock according to your ﬁnding root procedure and according to the result, repeat previous stage until the solution is obtained.

M M 4fL 4fL M M 1 2 D up D down x y 3.0000 1.0000 0.22019 0.57981 1.9899 0.57910

4fL (c) The way of the numerical procedure for solving this problem is by ﬁnding D that will up

produce M1 = 3. In the process Mx and My must be calculated (see the chapter on the program with its algorithms.).

9.7 Supersonic Branch In Chapter (8) it was shown that the isothermal model cannot describe adequately the situation because the thermal entry length is relatively large compared to the pipe length and the heat transfer is not sufficient to maintain constant temperature. In the Fanno model there is no heat transfer, and, furthermore, because the very limited amount of heat transformed it is closer to an adiabatic flow. The only limitation of the model is its uniform velocity (assuming parabolic flow for laminar and different profile for turbulent flow.). The information from the wall to the tube center6 is slower in reality. However, experiments from many starting with 1938 work by Frossel7 has shown that the error is not significant. Nevertheless, the comparison with reality shows that heat transfer cause changes to the flow and they need to be expected. These changes include the choking point at lower Mach number. To insert example on the change in the flow rate between isothermal flow to Fanno Flow. Insert also example on percentage of 9.8 Maximum length for the supersonic flow heat transfer. on the comparison of the maximum length of isother- It has to be noted and recognized that as opposed to subsonic branch the su- mal model and Fanno Model. personic branch has a limited length. It also must be recognized that there is a maximum length for which only supersonic flow can exist8. These results were obtained from the mathematical derivations but were verified by numerous exper- iments9. The maximum length of the supersonic can be evaluated when M = ∞ 6The word information referred to is the shear stress transformed from the wall to the center of the tube. 7See on the web http://naca.larc.nasa.gov/digidoc/report/tm/44/NACA-TM-844.PDF 8Many in the industry have difficulties in understanding this concept. The author seeks for a nice explanation of this concept for non–fluid mechanics engineers. This solicitation is about how to explain this issue to non-engineers or engineer without a proper background. 9If you have experiments demonstrating this point, please provide to the undersign so they can be added to this book. Many of the pictures in the literature carry copyright statements. 170 CHAPTER 9. FANNO FLOW as follows:

4fL 1 M 2 k + 1 k+1 M 2 max = − + ln 2 = D kM 2 2k k 1 2 2 1 + −2 M k + 1 (k + 1) 4fL (M ) −∞ + ln ∞ D → ∞ ∼ k 2k (k 1) × ∞ − ∞ 1 k + 1 (k + 1) = − + ln k 2k 2(k 1) − = 4fL (M , k = 1.4) = 0.8215 D → ∞ The maximum length of the supersonic ﬂow is limited by the above number. From the above analysis, it can be observed that no matter how high the entrance Mach number will be the tube length is limited and depends only on speciﬁc heat ratio, k as shown in Figure (9.5). The maximum length in supersonic flow

In Fanno Flow 1.5 1.4 1.3 1.2

max 1.1

D 1 4fL  0.9 0.8 0.7 0.6 0.5

maximum length, 0.4 0.3 0.2 0.1 0 1.2 1.25 1.3 1.35 1.4 1.45 1.5 1.55 1.6 1.65 Thu Mar 3 16:24:00 2005 spesific heat, k

Fig. 9.5: The maximum length as a function of speciﬁc heat, k

9.9 Working Conditions It has to be recognized that there are two regimes that can occur in Fanno flow model one of subsonic flow and the other supersonic flow. Even the flow in the tube starts as a supersonic in parts of the tube can be transformed into the subsonic 9.9. WORKING CONDITIONS 171 branch. A shock wave can occur and some portions of the tube will be in a subsonic flow pattern. The discussion has to differentiate between two ways of feeding the tube: converging nozzle or a converging-diverging nozzle. Three parameters, the dimensionless 4fL friction, D , the entrance Mach number, M1, and the pressure ratio, P2/P1 are controlling the flow. Only a combination of these two parameters is truly independent. However, all the three parameters can be varied and they are discussed separately here.

4fL

9.9.1 Variations of The Tube Length ( D ) Effects

MN U¨V

C8DFE

H¨I

N V

NOBP&Q8R4S V

:<;>=?

?@BA

J K

$&%

)+*-,/.102,43 5&687

' (

4fL Fig. 9.6: The effects of increase of D on the Fanno line

In the analysis of this effect, it should be assumed that back pressure is constant and/or low as possible as needed to maintain a choked ﬂow. First, the treatment of the two branches are separated.

Subsonic branch

For converging nozzle feeding, increasing the tube length results in increasing the exit Mach number (normally denoted herein as M2). Once the Mach number reaches maximum (M = 1), no further increase of the exit Mach number can be achieved. In this process, the mass ﬂow rate decreases. It is worth noting that entrance Mach number is reduced (as some might explain it to reduce the ﬂow rate). The entrance temperature increases as can be seen from Figure (9.7). The 172 CHAPTER 9. FANNO FLOW

constant pressure

lines

Y[Z \ 1’’ 2’’ 1’ 1 2’

Fanno lines ]

Fig. 9.7: The development properties in of converging nozzle

velocity therefore must decrease because the loss of the enthalpy (stagnation tem- P perature) is “used.” The density decrease because ρ = RT and when pressure is remains almost constant the density decreases. Thus, the mass flow rate must decrease. These results are applicable to the converging nozzle. In the case of the converging–diverging feeding nozzle, increase of the dimension- 4fL less friction, D , results in a similar flow pattern as in the converging nozzle. Once the flow becomes choked a different flow pattern emerges.

Supersonic Branch

There are several transitional points that change the pattern of the flow. Point a is the choking point (for the supersonic branch) in which the exit Mach number reaches to one. Point b is the maximum possible flow for supersonic flow and is not dependent on the nozzle. The next point, referred here as the critical point c, is the point in which no supersonic flow is possible in the tube i.e. the shock reaches to the nozzle. There is another point d, in which no supersonic flow is possible in the entire nozzle–tube system. Between these transitional points the effect parameters such as mass flow rate, entrance and exit Mach number are discussed. At the starting point the flow is choked in the nozzle, to achieve supersonic flow. The following ranges that has to be discussed includes (see Figure (9.8)): 9.9. WORKING CONDITIONS 173

0 < 4fL < 4fL 0 a D D choking → 4fL < 4fL < 4fL a b D choking D D shockless → 4fL < 4fL < 4fL b c D shockless D D chokeless → 4fL 4fL c D < D < chokeless ∞ → ∞

wyx

z+{

^`_ba |

all supersonic

flow b c

d j kmlonqpsrutqv mixed supersonic with subsonic flow with a shock the nozzle between is still

choked

}y~

efhg i

Fig. 9.8: The Mach numbers at entrance and exit of tube and mass ﬂow rate for Fanno Flow 4fL as a function of the D

The 0-a range, the mass flow rate is constant because the flow is choked at the nozzle. The entrance Mach number, M1 is constant because it is a function of the nozzle design only. The exit Mach number, M2 decreases (remember this flow is 4fL on the supersonic branch) and starts ( D = 0) as M2 = M1. At the end of the range a, M = 1. In the range of a b the flow is all supersonic. 2 − In the next range a b The flow is double choked and make the adjustment for − − the flow rate at different choking points by changing the shock location. The mass flow rate continues to be constant. The entrance Mach continues to be constant and exit Mach number is constant. b b 4fL The total maximum available for supersonic flow 0, D , is only a − − max theoretical length in which the supersonic flow can occur if nozzle is provided with a larger Mach number (a change to the nozzle area ratio which also reduces the mass flow rate). In the range b c, it is a more practical point. − In semi supersonic flow b c (in which no supersonic is available in the tube but − − only in the nozzle) the flow is still double choked and the mass flow rate is constant. Notice that exit Mach number, M2 is still one. However, the entrance Mach number, 4fL M1, reduces with the increase of D . It is worth noticing that in the a c the mass flow rate nozzle entrance velocity − − 174 CHAPTER 9. FANNO FLOW and the exit velocity remains constant!10 In the last range c the end is really the pressure limit or the break of the − −∞ model and the isothermal model is more appropriate to describe the flow. In this range, the flow rate decreases since (m˙ M )11. ∝ 1 To summarize the above discussion, Figures (9.8) exhibits the development of M1, 4fL M2 mass flow rate as a function of D . Somewhat different then the subsonic branch the mass flow rate is constant even if the flow in the tube is completely subsonic. This situation is because of the “double” choked condition in the nozzle. 4fL The exit Mach M2 is a continuous monotonic function that decreases with D . The entrance Mach M1 is a non continuous function with a jump at the point when shock occurs at the entrance “moves” into the nozzle. Fanno Flow M1 as a function of M2 1

4fL 0.9  = 0.1 D = 1.0 0.8 = 10.0 = 100.0 0.7

0.6

0.5

0.4 Entrace Mach number 0.3

0.2

0.1

0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Exit Mach number

Tue Oct 19 09:56:15 2004

4fL M1 M2 Fig. 9.9: as a function for various D

10On a personal note, this situation is rather strange to explain. On one hand, the resistance increases and on the other hand, the exit Mach number remains constant and equal to one. Does anyone have an explanation for this strange behavior suitable for non–engineers or engineers without background in ﬂuid mechanics? 11Note that ρ1 increases with decreases of M1 but this effect is less signiﬁcant. 9.9. WORKING CONDITIONS 175

Figure (9.9) exhibits the M1 as a function of M2. The Figure was calculated by 4fLmax utilizing the data from Figure (9.2) by obtaining the D for M2 and subtracting 4fL the given D and ﬁnding the corresponding M1. Fanno Flow

M1 as a function of M2 for the subsonic brench 5 4fL 4.5  = 0.1 D 4 = 0.2 = 0.4 3.5 = 0.1 shock = 0.4 3 1

M 2.5

1.5

0.5

0 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 M2

Tue Jan 4 11:26:19 2005

4fL M1 M2 Fig. 9.10: as a function for different D for supersonic entrance velocity.

The Figure (9.10) exhibits the entrance Mach number as a function of the M2. Obviously there can be two extreme possibilities for the subsonic exit branch. Sub- sonic velocity occurs for supersonic entrance velocity, one, when the shock wave occurs at the tube exit and two, at the tube entrance. In Figure (9.10) only for 4fL 4fL 4fL D = 0.1 and D = 0.4 two extremes are shown. For D = 0.2 shown with only 4fL shock at the exit only. Obviously, and as can be observed, the larger D creates larger differences between exit Mach number for the different shock locations. The 4fL larger D larger M1 must occurs even for shock at the entrance. 4fL For a given D , below the maximum critical length, the supersonic entrance ﬂow has three different regimes which depends on the back pressure. One, shockless ﬂow, tow, shock at the entrance, and three, shock at the exit. Below, the maximum 176 CHAPTER 9. FANNO FLOW critical length is mathematically

1 1 + k k + 1 4fL > + ln D −k 2k k 1 − 4fL For cases of D above the maximum critical length no supersonic flow can be over the whole tube and at some point a shock will occur and the flow becomes subsonic flow12.

P2 9.9.2 The Pressure Ratio, P1 , effects In this section the studied parameter is the variation of the back pressure and thus, the pressure ratio P2 variations. For very low pressure ratio the flow can be P1 assumed as incompressible with exit Mach number smaller than < 0.3. As the pressure ratio increases (smaller back pressure, P2), the exit and entrance Mach 4fL numbers increase. According to Fanno model the value of D is constant (friction factor, f, is independent of the parameters such as, Mach number, Reynolds number et cetera) thus the flow remains on the same Fanno line. For cases where the supply come from a reservoir with a constant pressure, the entrance pressure decreases as well because of the increase in the entrance Mach number (velocity). Again a differentiation of the feeding is important to point out. If the feeding nozzle is converging than the flow will be only subsonic. If the nozzle is “converging– diverging” than in some part supersonic flow is possible. At first the converging nozzle is presented and later the converging-diverging nozzle is explained.

Choking explanation for pressure variation/reduction

Decreasing the pressure ratio or in actuality the back pressure, results in increase of the entrance and the exit velocity until a maximum is reached for the exit velocity. The maximum velocity is when exit Mach number equals one. The Mach number, as it was shown in Chapter (4), can increases only if the area increase. In our model the tube area is postulated as a constant therefore the velocity cannot increase any further. However, for the flow to be continuous the pressure must decrease and for that the velocity must increase. Something must break since there are conflicting demands and it result in a “jump” in the flow. This jump is referred to as a choked flow. Any additional reduction in the back pressure will not change the situation in the tube. The only change will be at tube surroundings which are irrelevant to this discussion. If the feeding nozzle is a “converging–diverging” then it has to be differentiated 4fL between two cases; One case is where the D is short or equal to the critical 4fLmax length. The critical length is the maximum D that associate with entrance Mach number.

12See more on the discussion about changing the length of the tube.

9.9. WORKING CONDITIONS 177

a shock in the nozzle fully subsoinic

flow

critical Point a

criticalPoint b critical Point c q

critical Point d

4fL 4fL Fig. 9.11: The pressure distribution as a function of D for a short D

4fL Short D

Figure (9.12) shows different pressure profiles for different back pressures. Before the flow reaches critical point a (in the Figure) the flow is subsonic. Up to this stage the nozzle feeding the tube increases the mass flow rate (with decreasing back pressure). Between point a and point b the shock is in the nozzle. In this range and further reduction of the pressure the mass flow rate is constant no matter how low the back pressure is reduced. Once the back pressure is less than point b the supersonic reaches to the tube. Note however that exit Mach number, M2 < 1 and is not 1. A back pressure that is at the critical point c results in a shock wave that is at the exit. When the back pressure is below point c, the tube is “clean” of any shock13. The back pressure below point c has some adjustment as it occurs with exceptions of point d.

13It is common misconception that the back pressure has to be at point d.

178 CHAPTER 9. FANNO FLOW

Ð&ÑÓÒsÔ ÐÖÕsÐØ×"Ù·Ô Ú¤Ô ×"ÑÜÛÞÝßà·áâ¦ã

ämå æ

a shock in the nozzle

fully subsoinic

¤B¥¤¦§©¨¢ª¬«s¯®±°³²´µ·¶q¸¹

flows ¢¡£

ºm» ¼

{

¾À¿ÂÁÄÃ

Å³ÆÇÈ·ÉÊ¦Ë

ÌmÍ Î

critical Point a

criticalPoint b critical Point c

4fL 4fL Fig. 9.12: The pressure distribution as a function of D for a long D

4fL Long D

4fL 4fLmax In the case of D > D reduction of the back pressure results in the same 4fL process as explained in the short D up to point c. However, point c in this case is 4fL 4fLmax different from point c at the case of short tube D < D . In this point the exit Mach number is equal to 1 and the ﬂow is double shock. Further reduction of the back pressure at this stage will not “move” the shock wave downstream the nozzle. At point c or location of the shock wave, is a function entrance Mach number, M1 4fL and the “extra” D . The is no analytical solution for the location of this point c. The procedure is (will be) presented in later stage.

9.9.3 Entrance Mach number, M1, effects

In this discussion, the effect of changing the throat area on the nozzle efficiency is neglected. In reality these effects have significance and needs to be accounted for some instances. This dissection deals only with the flow when it reaches the 9.9. WORKING CONDITIONS 179 Mach number in Fanno Flow

4fL  D 2

1.8

1.6 shock at 1.4 75% 1.2 50% 5% 1

0.8 Mach Number

0.6

0.4

0.2

0 0 0.05 0.1 0.15 0.2 0.25 4fL  D Tue Jan 4 12:11:20 2005

4fL Fig. 9.13: The effects of pressure variations on Mach number proﬁle as a function of D 4fL = 0.3 when the total resistance D for Fanno Flow

supersonic branch reached otherwise the ﬂow is subsonic with regular effects. It is assumed that in this discussion that the pressure ratio P2 is large enough to create P1 4fL a choked ﬂow and D is small enough to allow it to happen.

The entrance Mach number, M1 is a function of the ratio of the nozzle’s throat area to the nozzle exit area and its efficiency. This effect is the third parameter discussed here. Practically, the nozzle area ratio is changed by changing the throat area. 4fL As was shown before, there are two different maximums for D ; first is the total 4fL maximum D of the supersonic which depends only on the specific heat, k, and second the maximum depends on the entrance Mach number, M1. This analysis 4fL 4fLmax deals with the case where D is shorter than total D .

4fL 4fLmax Obviously, in this situation, the critical point is where D is equal to D as a result in the entrance Mach number. The process of decreasing the converging–diverging nozzle’s throat increases the 180 CHAPTER 9. FANNO FLOW P2/P1 Fanno Flow

4fL  D 4.8 4.4 4 3.6 3.2 5 % 50 % 2.8 75 % 2.4 P2/P1 2 1.6 1.2 0.8 0.4 0 0 0.05 0.1 0.15 0.2 0.25 4fL  D Fri Nov 12 04:07:34 2004

4fL 4fL = 0.3 Fig. 9.14: Fanno Flow Mach number as a function of D when the total D entrance14 Mach number. If the tube contains no supersonic flow then reducing the nozzle throat area wouldn’t increase the entrance Mach number. This part is for the case where some part of the tube is under supersonic regime and there is shock as a transition to subsonic branch. Decreasing the nozzle throat area moves the shock location downstream. The “payment” for increase in the supersonic length is by reducing the mass flow. Further, decrease of the throat area results in flushing the shock out of the tube. By doing so, the throat area decreases. The mass flow rate is proportionally linear to the throat area and therefore the mass flow rate reduces. The process of decreasing the throat area also results in increasing the pressure drop of the nozzle (larger resistance in the nozzle15)16. 4fL 4fLmax In the case of large tube D > D the exit Mach number increases with the

14The word “entrance” referred to the tube and not to the nozzle. The reference to the tube is because it is the focus of the study. 15Strange? Frictionless nozzle has a larger resistance when the throat area decreases 16It is one of the strange phenomenon that in one way increasing the resistance (changing the throat 4fL area) decreases the ﬂow rate while in a different way (increasing the D ) does not affect the ﬂow rate.

9.9. WORKING CONDITIONS 181

üþý ÿ

¢ £¥¤§¦©¨

"!# %$ !

öø÷ ùûú

èêé<ë ìêíïîñðò8óô¦õõ

shock

Fig. 9.15: Schematic of a “long” tube in supersonic branch decrease of the throat area. Once the exit Mach number reaches one no further increases is possible. However, the location of the shock wave approaches to the theoretical location if entrance Mach, M = .

1 ∞

? @

57689;:(<*=

SUTVXW

YUZ[]\ LNMPORQ

B(C*D &('*)

, F

+ E

, F

, -/.102-3.%4/0 F GIHKJ

4fL Fig. 9.16: The extra tube length as a function of the shock location, D supersonic branch

The maximum location of the shock The main point in this discussion however, is to ﬁnd the furtherest shock location downstream. Figure (9.16) shows the 4fL possible ∆ D as function of retreat of the location of the shock wave from the maximum location. When the entrance Mach number is inﬁnity, M1 = , if the ∞ 182 CHAPTER 9. FANNO FLOW shock location is at the maximum length, then shock at Mx = 1 results in My = 1. The proposed procedure is based on Figure (9.16).

4fL 4fL i) Calculate the extra D and subtract the actual extra D assuming shock at the left side (at the max length).

4fL 4fL ii) Calculate the extra D and subtract the actual extra D assuming shock at the right side (at the entrance).

iii) According to the positive or negative utilizes your root ﬁnding procedure. cRd3ePf g

1 ^`_Ia

4 b fL

D max∞

4fL M1 Fig. 9.17: The maximum entrance Mach number, to the tube as a function of D supersonic branch

From numerical point of view, the Mach number equal infinity when left side as- sumes result in infinity length of possible extra (the whole flow in the tube is subsonic). To overcome this numerical problem it is suggested to start the calculation from distance from the right hand side. Let denote

4fL 4f¯L 4fL ∆ = (9.51) D D actual − D sup 4fL 4fL Note that D is smaller than D . The requirement that has to be sat- sup max ∞ 4fL isﬁed is that denote D as diff erence between the maximum possible of retreat length in which the supersonic ﬂow is achieved and the actual length in which the

9.9. WORKING CONDITIONS 183 ﬂow is supersonic see Figure (9.15). The retreating length is expressed as subsonic but

4fL 4fL 4fL D = D D (9.52) retreat max − sup ∞

Figure (9.17) shows the entrance Mach number, M1 reduces after the maximum length is exceeded.

Example 9.3: 4fL Calculate the shock location for entrance Mach number M1 = 8 and for D = 0.9 assume that k = 1.4 (Mexit = 1).

SOLUTION 4fLmax The solution is obtained by an iterative process. The maximum D for k = 1.4 4fL 4fL is 0.821508116. Hence, D exceed the maximum length D for this entrance 4fL Mach number. The maximum for M1 = 8 is D = 0.76820, thus the extra tube is ∆ 4fL = 0.9 0.76820 = 0.1318. The left side is when the shock occurs at D − 4fL 4fL D = 0.76820 (ﬂow is choked and no additional D ). Hence, the value of left side is 0.1318. The right side is when the shock is at the entrance at which the extra 4fL− D is calculated for Mx and My is

Ty ρy Py P0y Mx My Tx ρx Px P0x 8.0000 0.39289 13.3867 5.5652 74.5000 0.00849

With (M1)0

M 4fL P P0 ρ U T D P∗ P0∗ ρ∗ U∗ T∗ 0.39289 2.4417 2.7461 1.6136 2.3591 0.42390 1.1641

The extra ∆ 4fL is 2.442 0.1318 = 2.3102 Now the solution is somewhere D − between the negativ e of left side to the positive of the right side17. In a summary of the actions is done by the following algorithm:

4fL 4fL (a) check if the D exceeds the maximum D max for the supersonic ﬂow. Accordingly continue. (b) Guess 4fL = 4fL 4fL D up D − D max 4fL (c) Calculate the Mach number corresponding to the current guess of D up,

17What if the right side is also negative? The ﬂow is chocked and shock must occur in the nozzle before entering the tube. Or in a very long tube the whole ﬂow will be subsonic. 184 CHAPTER 9. FANNO FLOW

(d) Calculate the associate Mach number, Mx with the Mach number, My calculated previously,

4fL (e) Calculate D for supersonic branch for the Mx

4fL (f) Calculate the “new and improved” D up

(g) Compute the “new 4fL = 4fL 4fL D down D − D up

4fL (h) Check the new and improved D down against the old one. If it is satisfactory stop or return to stage (b).

Shock location are:

M M 4fL 4fL M M 1 2 D up D down x y 8.0000 1.0000 0.57068 0.32932 1.6706 0.64830

The iteration summary is also shown below

i 4fL 4fL M M 4fL D up D down x y D 0 0.67426 0.22574 1.3838 0.74664 0.90000 1 0.62170 0.27830 1.5286 0.69119 0.90000 2 0.59506 0.30494 1.6021 0.66779 0.90000 3 0.58217 0.31783 1.6382 0.65728 0.90000 4 0.57605 0.32395 1.6554 0.65246 0.90000 5 0.57318 0.32682 1.6635 0.65023 0.90000 6 0.57184 0.32816 1.6673 0.64920 0.90000 7 0.57122 0.32878 1.6691 0.64872 0.90000 8 0.57093 0.32907 1.6699 0.64850 0.90000 9 0.57079 0.32921 1.6703 0.64839 0.90000 10 0.57073 0.32927 1.6705 0.64834 0.90000 11 0.57070 0.32930 1.6706 0.64832 0.90000 12 0.57069 0.32931 1.6706 0.64831 0.90000 13 0.57068 0.32932 1.6706 0.64831 0.90000 14 0.57068 0.32932 1.6706 0.64830 0.90000 15 0.57068 0.32932 1.6706 0.64830 0.90000 16 0.57068 0.32932 1.6706 0.64830 0.90000 17 0.57068 0.32932 1.6706 0.64830 0.90000

This procedure rapidly converted to the solution. 9.10. THE APPROXIMATION OF THE FANNO FLOW BY ISOTHERMAL FLOW185 9.10 The Approximation of the Fanno flow by Isothermal Flow The isothermal flow model has an equation that theoreticians find it easier to use compared to the Fanno flow model.

One must notice that the maximum temperature at the entrance is T01. When the Mach number decreases the temperature approaches the stagnation temperature (T T ). Hence, if one allows certain deviation of temperature, say about 1% that → 0 flow can be assumed to be isothermal. This tolerance requires that (T T )/T = 0 − 0 0.99 which requires that enough for M1 < 0.15 even for large k = 1.67. This 4fL requirement provides that somewhere (depend) in the vicinity of D = 25 the 4fL flow can be assumed isothermal. Hence the mass flow rate is a function of D because M1 changes. Looking at the table or Figure (9.2) or the results from computer program attached to this book shows that reduction of the mass flow is very rapid. to insert a question or example about this issue in end

M1 Fanno flow with comperison to Isothermal Flow

0.4

P2 / P1 = 0.1 iso

P2 / P1 = 0.8 iso 0.3 P2 / P1 = 0.1

P2 / P1 = 0.2 1 P / P = 0.5 M 2 1 0.2 P2 / P1 = 0.8

0.1

0 0 10 20 30 40 50 60 70 80 90 100 4fL  Wed Mar 9 11:38:27 2005 D

Fig. 9.18: The entrance Mach number as a function of dimensionless resistance and comparison with Isothermal Flow

4fL As it can be seen for the Figure (9.18) the dominating parameter is D . The results are very similar for isothermal ﬂow. The only difference is in small dimensionless 4fL friction, D . 186 CHAPTER 9. FANNO FLOW 9.11 More Examples of Fanno Flow Example 9.4: To demonstrate the utility in Figure (9.18) consider the following example. Find the mass ﬂow rate for f = 0.05, L = 4[m], D = 0.02[m] and pressure ratio P2/P1 = 0.1, 0.3, 0.5, 0.8. The stagnation conditions at the entrance are 300K and 3[bar] air.

SOLUTION 4fL First calculate the dimensionless resistance, D .

4 0.05 4 4fL = × × = 40 D 0.02

From Figure (9.18) for P /P = 0.1 M 0.13 etc. 2 1 1 ≈ or accurately by utilizing the program as in the following table.

4fL 4fL 4fL P2 M1 M2 D D 1 D 2 P1 0.12728 1.0000 40.0000 40.0000 0.0 0.11637 0.12420 0.40790 40.0000 42.1697 2.1697 0.30000 0.11392 0.22697 40.0000 50.7569 10.7569 0.50000 0.07975 0.09965 40.0000 107.42 67.4206 0.80000

Only for the pressure ratio of 0.1 the ﬂow is choked.

M T ρ A P A P ? × T0 ρ0 A P0 A P0 ∗× 0.12728 0.99677 0.99195 4.5910 0.98874 4.5393 0.12420 0.99692 0.99233 4.7027 0.98928 4.6523 0.11392 0.99741 0.99354 5.1196 0.99097 5.0733 0.07975 0.99873 0.99683 7.2842 0.99556 7.2519

Therefore, T T and is the same for the pressure. Hence, the mass rate is ≈ 0 a function of the Mach number. The Mach number is indeed a function of the pressure ratio but mass ﬂow rate is a function of pressure ratio only through Mach number. The mass ﬂow rate is

k π 0.022 1.4 kg m˙ = P AM = 300000 × 0.127 0.48 RT × 4 × × 287300 ≈ sec r r 9.12. THE TABLE FOR FANNO FLOW 187 and for the rest P 0.1242 kg m˙ 2 = 0.3 0.48 = 0.468 P ∼ × 0.1273 sec 1 P 0.1139 kg m˙ 2 = 0.5 0.48 = 0.43 P ∼ × 0.1273 sec 1 P 0.07975 kg m˙ 2 = 0.8 0.48 = 0.30 P ∼ × 0.1273 sec 1

9.12 The Table for Fanno Flow

Table 9.1: Fanno Flow Standard basic Table

M 4fL P P0 ρ U T D P∗ P0∗ ρ∗ U∗ T∗ 0.03 787.08 36.5116 19.3005 30.4318 0.03286 1.1998 0.04 440.35 27.3817 14.4815 22.8254 0.04381 1.1996 0.05 280.02 21.9034 11.5914 18.2620 0.05476 1.1994 0.06 193.03 18.2508 9.6659 15.2200 0.06570 1.1991 0.07 140.66 15.6416 8.2915 13.0474 0.07664 1.1988 0.08 106.72 13.6843 7.2616 11.4182 0.08758 1.1985 0.09 83.4961 12.1618 6.4613 10.1512 0.09851 1.1981 0.10 66.9216 10.9435 5.8218 9.1378 0.10944 1.1976 0.20 14.5333 5.4554 2.9635 4.5826 0.21822 1.1905 0.25 8.4834 4.3546 2.4027 3.6742 0.27217 1.1852 0.30 5.2993 3.6191 2.0351 3.0702 0.32572 1.1788 0.35 3.4525 3.0922 1.7780 2.6400 0.37879 1.1713 0.40 2.3085 2.6958 1.5901 2.3184 0.43133 1.1628 0.45 1.5664 2.3865 1.4487 2.0693 0.48326 1.1533 0.50 1.0691 2.1381 1.3398 1.8708 0.53452 1.1429 0.55 0.72805 1.9341 1.2549 1.7092 0.58506 1.1315 0.60 0.49082 1.7634 1.1882 1.5753 0.63481 1.1194 0.65 0.32459 1.6183 1.1356 1.4626 0.68374 1.1065 0.70 0.20814 1.4935 1.0944 1.3665 0.73179 1.0929 0.75 0.12728 1.3848 1.0624 1.2838 0.77894 1.0787 0.80 0.07229 1.2893 1.0382 1.2119 0.82514 1.0638 0.85 0.03633 1.2047 1.0207 1.1489 0.87037 1.0485 0.90 0.01451 1.1291 1.0089 1.0934 0.91460 1.0327 0.95 0.00328 1.061 1.002 1.044 0.95781 1.017 1.00 0.0 1.00000 1.000 1.000 1.00 1.000 2.00 0.30500 0.40825 1.688 0.61237 1.633 0.66667 3.00 0.52216 0.21822 4.235 0.50918 1.964 0.42857 188 CHAPTER 9. FANNO FLOW Table 9.1: Fanno Flow Standard basic Table (continue)

M 4fL P P0 ρ U T D P∗ P0∗ ρ∗ U∗ T∗ 4.00 0.63306 0.13363 10.72 0.46771 2.138 0.28571 5.00 0.69380 0.089443 25.00 0.44721 2.236 0.20000 6.00 0.72988 0.063758 53.18 0.43568 2.295 0.14634 7.00 0.75280 0.047619 1.0E+2 0.42857 2.333 0.11111 8.00 0.76819 0.036860 1.9E+2 0.42390 2.359 0.086957 9.00 0.77899 0.029348 3.3E+2 0.42066 2.377 0.069767 10.00 0.78683 0.023905 5.4E+2 0.41833 2.390 0.057143 20.00 0.81265 0.00609 1.5E+4 0.41079 2.434 0.014815 25.00 0.81582 0.00390 4.6E+4 0.40988 2.440 0.00952 30.00 0.81755 0.00271 1.1E+5 0.40938 2.443 0.00663 35.00 0.81860 0.00200 2.5E+5 0.40908 2.445 0.00488 40.00 0.81928 0.00153 4.8E+5 0.40889 2.446 0.00374 45.00 0.81975 0.00121 8.6E+5 0.40875 2.446 0.00296 50.00 0.82008 0.000979 1.5E+6 0.40866 2.447 0.00240 55.00 0.82033 0.000809 2.3E+6 0.40859 2.447 0.00198 60.00 0.82052 0.000680 3.6E+6 0.40853 2.448 0.00166 65.00 0.82066 0.000579 5.4E+6 0.40849 2.448 0.00142 70.00 0.82078 0.000500 7.8E+6 0.40846 2.448 0.00122 CHAPTER 10

RAYLEIGH FLOW

Rayleigh flow is (frictionless) flow with heat transfer through a pipe of constant cross sectional area. In practice Rayleigh flow is really provide good model for the real situation. Yet, Rayleigh flow is practical and useful concept in a obtaining trends and limits. The density and pressure change due to external cooling or heating. As opposed to the two previous models, the heat transfer can be in two directions not like the friction (there is no negative friction). This fact create situation different compare to the previous two models. This model applied to case where the heat transfer is significant and the friction can be ignored.

10.1 Introduction

The third simple model for an one dimen- hji flow

direction prq mon

sional flow is for constant heat transfer for fric- k`l uwv tionless flow. This flow referred in the literature s(t as Rayleigh Flow (see historical notes). This flow is another extreme case in which the fric- x tion effect are neglected because their relative heat transfer (in and out) effect is much smaller the heat transfer effect. While the isothermal flow model has heat trans- Fig. 10.1: The control volume of fer and friction the main assumption was that rel- Rayleigh Flow ative length is so the heat transfer occurs between the surrounding and tube. In contrast, the heat transfer in Rayleigh flow occurs either between unknown temperature to tube and the heat flux is maintained constant. As before, a simple model is built around assumption of constant properties (poorer prediction to case were chemical reaction take palace). This model usage is to have a rough predict the conditions occur mostly in

189 190 CHAPTER 10. RAYLEIGH FLOW situations involve chemical reaction. In analysis of the flow, one has to be aware that properties do change significantly for a large range of temperature. Yet, for smaller range of temperature and length the calculations are more accurate. Nev- ertheless, the main characteristic of the flow such as chocking condition etc. are encapsulated in this model. The basic physics of the flow revolves around the fact that the gas is highly compressible. The density change though the heat transfer (temperature change). As appose to Fanno flow in which the resistance always oppose the the flow direction, in Rayleigh flow also cooling can be applied. The flow velocity acceleration change the direction when the cooling is applied.

10.2 Governing Equation The energy balance on the control volume reads Q = C (T T ) (10.1) p 02 − 01 the momentum balance reads A(P P ) = m˙ (V V ) (10.2) 1 − 2 2 − 1 The mass conservation reads

ρ1U1A = ρ2U2A = m˙ (10.3) Equation of state P P 1 = 2 (10.4) ρ1T1 ρ2T2 There are four equations with four unknown, if the upstream conditions are known (or downstream condition are known). Thus, a solution can be obtained. One can notice that equations (10.2), (10.3) and (10.4) are similar to the equations that were solved for the shock wave. 2 P2 1 + kM1 = 2 (10.5) P1 1 + kM2 The equation of state (10.4) can further assist in obtaining the temperature ratio as T P ρ 2 = 2 1 (10.6) T1 P1 ρ2 The density ratio can be expressed in term of mass conservation as

U2 √kRT2 kRT ρ1 U2 2 M2 T2 = = r = (10.7) ρ2 U1 U1 M1 T1 kRT1 r kRT1 r q 10.2. GOVERNING EQUATION 191 Substituting equations (10.5) and (10.7) into equation (10.6) yields

T 1 + kM 2 M T 2 = 1 2 2 (10.8) T 2 M T 1 1 + kM2 1 r 1 Transferring the temperature ratio to left hand side and squaring results in

2 2 2 T2 1 + kM1 M2 = 2 (10.9) T1 1 + kM M1

{N|~}

N X

constant Pressure line z

Fig. 10.2: The Temperature Entropy Diagram For Rayleigh Line

The Rayleigh line exhibits two possible maximums one for dT/ds = 0 and for ds/dT = 0. The second maximum can be expressed as dT/ds = The second ∞ law is used to ﬁnd the expression for derivative. s s T k 1 P 1 − 2 = ln 2 − ln 2 (10.10) Cp T1 − k P1

2 2 s1 s2 1 + kM1 ) M2 k 1 1 + kM21 − = 2 ln ( 2 + − ln 2 (10.11) Cp (1 + kM ) M1 k 1 + kM 2 1

Let the initial condition M1, and s1 are constant then the variable parameters are M2, and s2. A derivative of equation (10.11) results in 1 ds 2(1 M 2) = − 2 (10.12) Cp dM M(1 + kM ) 192 CHAPTER 10. RAYLEIGH FLOW Take the derivative of the equation (10.12) when letting the variable parameters be T2, and M2 results in

dT 1 kM 2 = constant − (10.13) dM × (1 + kM 2)3

Combining equations (10.12) and (10.13) by eliminating dM results in

dT M(1 kM 2) = constant − (10.14) ds × (1 M 2)(1 + kM 2)2 − On T-s diagram a family of curves can be drawn for a given constant. Yet for every curve, several observations can be generalized. The derivative is equal to zero when 1 kM 2 = 0 or M = 1/√k or when M 0. The derivative is equal to infinity, − → dT/ds = when M = 1. From thermodynamics, increase of heating results in ∞ increase of entropy. And cooling results in reduction of entropy. Hence, when cooling applied to a tube the velocity decreases and heating applied the velocity increases. The peculiars point of M = 1/√k when additional heat is applied the temperature is decreasing. The derivative is negative, dT/ds < 0, yet note this point is not the choking point. The chocking is occurred only when M = 1 because it violate the second law. The transition to supper sonic flow occurs when the area changes, some what similarly to Fanno flow, Yet, chocking can be explained by the fact increase of energy must accompanied by increase of entropy. But the entropy of supersonic flow is lower (see the Figure (10.2)) and therefore it is not possible (the maximum entropy at M = 1.). It is convent to referrers to the value of M = 1. These value referred as the “star”1 values. The equation (10.5) can be written between chocking point and any point on the curve.

2 P ∗ 1 + kM1 = (10.15) P1 1 + k

The temperature ratio is

2 2 T ∗ 1 1 + kM = 1 (10.16) T M 2 1 + k 1

U ∗ √ ρ1 U ∗ √ kRT ∗ 1 T = = kRT ∗ = ∗ (10.17) U1 ρ∗ U1 √kRT1 M1 T1 √kRT1 r

1The star is an asterisk. 10.3. RAYLEIGH FLOW TABLES 193

k 1 2 2 T01 T1 1 + −2 M1 2(1 + k)M1 k 1 2 = 1+k = 2 2 1 + − M1 (10.18) T0∗ T (1 + kM ) 2 ∗ 2 The stagnation pressure ratio reads

k k 1 k 1 2 2 − P01 P1 1 + −2 M1 1 + k 1 + kM1 = 1+k = 2 (1+k) (10.19) P0∗ P 1 + kM ∗ 2 1 2 ! 10.3 Rayleigh Flow Tables The “star” values are tabulated in Table (10.1). Several observations can be made in regards to the stagnation temperature.

Table 10.1: Rayleigh Flow k=1.4

M T T0 P P0 ρ∗ T∗ T0∗ P∗ P0∗ ρ 0.03 0.00517 0.00431 2.397 1.267 0.00216 0.04 0.00917 0.00765 2.395 1.266 0.00383 0.05 0.014300 0.011922 2.392 1.266 0.00598 0.06 0.020529 0.017119 2.388 1.265 0.00860 0.07 0.027841 0.023223 2.384 1.264 0.011680 0.08 0.036212 0.030215 2.379 1.262 0.015224 0.09 0.045616 0.038075 2.373 1.261 0.019222 0.10 0.056020 0.046777 2.367 1.259 0.023669 0.20 0.20661 0.17355 2.273 1.235 0.090909 0.25 0.30440 0.25684 2.207 1.218 0.13793 0.30 0.40887 0.34686 2.131 1.199 0.19183 0.35 0.51413 0.43894 2.049 1.178 0.25096 0.40 0.61515 0.52903 1.961 1.157 0.31373 0.45 0.70804 0.61393 1.870 1.135 0.37865 0.50 0.79012 0.69136 1.778 1.114 0.44444 0.55 0.85987 0.75991 1.686 1.094 0.51001 0.60 0.91670 0.81892 1.596 1.075 0.57447 0.65 0.96081 0.86833 1.508 1.058 0.63713 0.70 0.99290 0.90850 1.423 1.043 0.69751 0.75 1.014 0.94009 1.343 1.030 0.75524 0.80 1.025 0.96395 1.266 1.019 0.81013 0.85 1.029 0.98097 1.193 1.011 0.86204 0.90 1.025 0.99207 1.125 1.005 0.91097 0.95 1.015 0.99814 1.060 1.001 0.95693 194 CHAPTER 10. RAYLEIGH FLOW Table 10.1: Rayleigh Flow k=1.4 (continue)

M T T0 P P0 ρ∗ T∗ T0∗ P∗ P0∗ ρ 1.0 1.00 1.00 1.00 1.00 1.000 1.1 0.96031 0.99392 0.89087 1.005 1.078 1.2 0.91185 0.97872 0.79576 1.019 1.146 1.3 0.85917 0.95798 0.71301 1.044 1.205 1.4 0.80539 0.93425 0.64103 1.078 1.256 1.5 0.75250 0.90928 0.57831 1.122 1.301 1.6 0.70174 0.88419 0.52356 1.176 1.340 1.7 0.65377 0.85971 0.47562 1.240 1.375 1.8 0.60894 0.83628 0.43353 1.316 1.405 1.9 0.56734 0.81414 0.39643 1.403 1.431 2.0 0.52893 0.79339 0.36364 1.503 1.455 2.1 0.49356 0.77406 0.33454 1.616 1.475 2.2 0.46106 0.75613 0.30864 1.743 1.494 2.3 0.43122 0.73954 0.28551 1.886 1.510 2.4 0.40384 0.72421 0.26478 2.045 1.525 2.5 0.37870 0.71006 0.24615 2.222 1.538 2.6 0.35561 0.69700 0.22936 2.418 1.550 2.7 0.33439 0.68494 0.21417 2.634 1.561 2.8 0.31486 0.67380 0.20040 2.873 1.571 2.9 0.29687 0.66350 0.18788 3.136 1.580 3.0 0.28028 0.65398 0.17647 3.424 1.588 3.5 0.21419 0.61580 0.13223 5.328 1.620 4.0 0.16831 0.58909 0.10256 8.227 1.641 4.5 0.13540 0.56982 0.081772 12.50 1.656 5.0 0.11111 0.55556 0.066667 18.63 1.667 5.5 0.092719 0.54473 0.055363 27.21 1.675 6.0 0.078487 0.53633 0.046693 38.95 1.681 6.5 0.067263 0.52970 0.039900 54.68 1.686 7.0 0.058264 0.52438 0.034483 75.41 1.690 7.5 0.050943 0.52004 0.030094 1.0E+2 1.693 8.0 0.044910 0.51647 0.026490 1.4E+2 1.695 8.5 0.039883 0.51349 0.023495 1.8E+2 1.698 9.0 0.035650 0.51098 0.020979 2.3E+2 1.699 9.5 0.032053 0.50885 0.018846 3.0E+2 1.701 10.0 0.028972 0.50702 0.017021 3.8E+2 1.702 20.0 0.00732 0.49415 0.00428 1.1E+4 1.711 25.0 0.00469 0.49259 0.00274 3.2E+4 1.712 30.0 0.00326 0.49174 0.00190 8.0E+4 1.713 35.0 0.00240 0.49122 0.00140 1.7E+5 1.713 10.3. RAYLEIGH FLOW TABLES 195 Table 10.1: Rayleigh Flow k=1.4 (continue)

M T T0 P P0 ρ∗ T∗ T0∗ P∗ P0∗ ρ 40.0 0.00184 0.49089 0.00107 3.4E+5 1.714 45.0 0.00145 0.49066 0.000846 6.0E+5 1.714 50.0 0.00117 0.49050 0.000686 1.0E+6 1.714 55.0 0.000971 0.49037 0.000567 1.6E+6 1.714 60.0 0.000816 0.49028 0.000476 2.5E+6 1.714 65.0 0.000695 0.49021 0.000406 3.8E+6 1.714 70.0 0.000600 0.49015 0.000350 5.5E+6 1.714

The Data is presented in Figure (10.3).

Rayleigh Flow k = 1 4 4 * T / T * T0 / T0 * 3 P / P * P0 / P0 ∗ ρ / ρ 2 @( M = ∞ ) 1.7143

1 @( M = ∞ ) 0.4898

0 0.1 1 10 M

Fri May 6 11:11:44 2005

Fig. 10.3: The basic functions of Rayleigh Flow (k=1.4) 196 CHAPTER 10. RAYLEIGH FLOW 10.4 Examples For Rayleigh Flow Illustrative example The typical questions that raised in Rayleigh Flow are related to the maximum heat that can be transfered to gas (reaction heat) and to ﬂow rate.

Example 10.1: Air enters a pipe with pressure of 3[bar] and temperature of 27◦C at Mach number of M = 0.25. Due internal combustion heat was released and the exit temperature was found to be 127◦C. Calculated the exit Mach number, the exit pressure, the total exit pressure, and heat released (transfered) to the air. After what amount of kJ energy the exit temperature will start to decrease? Assume CP = 1.004 kg◦C h i SOLUTION The entrance Mach number and the exit temperature are given and from the Table 10.1 or from the program the initial ratio can be calculated. From the initial values the ratio at the exit can be computed as following.

M T T0 P P0 ρ∗ T∗ T0∗ P∗ P0∗ ρ 0.25000 0.30440 0.25684 2.2069 1.2177 0.13793 and T T T 400 2 = 1 2 = 0.304 = 0.4053 T ∗ T ∗ T1 × 300

M T T0 P P0 ρ∗ T∗ T0∗ P∗ P0∗ ρ 0.29831 0.40530 0.34376 2.1341 1.1992 0.18991

The exit Mach number is known, the exit pressure can be calculated as

P ∗ P2 1 P2 = P1 = 3 2.1341 = 2.901[Bar] P1 P ∗ × 2.2069 × For the entrance the stagnation values are

M T ρ A P A P F ? × T0 ρ0 A P0 A P0 F ∗× ∗ 0.25000 0.98765 0.96942 2.4027 0.95745 2.3005 1.0424

The total exit pressure, P02 can be calculated as the following:

isentropic

P01 P0∗ P02 1 1 P02 = P1 = 3 1.1992 = 3.08572[Bar] z}|{P1 P01 P0∗ × 0.95745 × 1.2177 × 10.4. EXAMPLES FOR RAYLEIGH FLOW 197 The heat release (heat transfer) can be calculated from obtaining the stagnation temperature form the both sides. The stagnation temperature at the entrance, T01 isentropic

T01 T01 = T1 = 300/0.98765 = 303.75[K] z}|{T1 The isentropic conditions at the exit are

M T ρ A P A P F ? × T0 ρ0 A P0 A P0 F ∗× ∗ 0.29831 0.98251 0.95686 2.0454 0.94012 1.9229 0.90103

The exit stagnation temperature is

isentropic

T02 T02 = T2 = 400/0.98765 = 407.12[K] z}|{T2 The heat release becomes Q kJ = C (T T ) 1 1.004 (407.12 303.75) = 103.78 m˙ p 02 − 01 × × − seckg C ◦ The maximum temperature occurs at the point where the Mach number reaches 1/√k and at this point the Rayleigh relationship are:

M T T0 P P0 ρ∗ T∗ T0∗ P∗ P0∗ ρ 0.84515 1.0286 0.97959 1.2000 1.0116 0.85714

The maximum heat before the temperature can be calculated as following:

T ∗ Tmax 300 Tmax = T1 1.0286 = 1013.7[K] T1 T ∗ 0.3044 × The isentropic relationship at the maximum are

M T ρ A P A P F ? × T0 ρ0 A P0 A P0 F ∗× ∗ 0.84515 0.87500 0.71618 1.0221 0.62666 0.64051 0.53376

The stagnation temperature for this point is

T0max 1013.7 T0max = Tmax = = 1158.51[K] ∗ Tmax 0.875 The maximum heat can be calculated as Q kJ = C (T T ) = 1 1.004 (1158.51 303.75) = 858.18 m˙ p 0max − 01 × × − kgsecK Note that this point isn’t the choking point. 198 CHAPTER 10. RAYLEIGH FLOW Example 10.2: Heat is added to the air until the ﬂow is choked in amount of 600 [kJ/kg]. The exit temperature is 1000 [K]. Calculated the entrance temperature and the entrance Mach number.

SOLUTION The solution involve ﬁnding the stagnation temperature at the exit and subtraction of the heat (heat equation) to obtain the entrance stagnation temperature. From the Table (10.1) or from the Potto-GDC the following ratios can be obtained.

M T ρ A P A P F ? × T0 ρ0 A P0 A P0 F ∗× ∗ 1.0000 0.83333 0.63394 1.0000 0.52828 0.52828 0.52828

The stagnation temperature

T02 1000 T02 = T2 = = 1200.0[K] T2 0.83333 The entrance temperature is

T0 Q/m˙ 600 1 = 1 = 1200 = 0.5016 T − T C − 1200 1.004 ∼ 02 02 P × T01 It must be noted that T0 = T0∗. Therefore with = 0.5016 either by Table (10.1) 2 T0∗ or by Potto-GDC the following is obtained

M T T0 P P0 ρ∗ T∗ T0∗ P∗ P0∗ ρ 0.34398 0.50160 0.42789 2.0589 1.1805 0.24362

Thus, entrance Mach number is 0.38454 and the entrance temperature can be calculated as following

T1 T1 = T ∗ = 1000 0.58463 = 584.6[K] T ∗ × The difference between the supersonic branch to subsonic branch

Example 10.3: Air with Mach 3 enters a frictionless duct with heating. What is the maximum heat that can be add so there is no subsonic ﬂow. If a shock is occurs immediately at the entrance what is the maximum heat that can be added?

SOLUTION To achieve maximum heat transfer the exit Mach number has to be one, M2 = 1.

Q T01 = Cp (T02 T01 ) = CpT0∗ 1 m˙ − − T ∗ 0 The table for M = 3 as following 10.4. EXAMPLES FOR RAYLEIGH FLOW 199

M T T0 P P0 ρ∗ T∗ T0∗ P∗ P0∗ ρ 3.0000 0.28028 0.65398 0.17647 3.4245 1.5882

The higher the entrance stagnation temperature the larger the heat amount that can be absorbed by the ﬂow. In subsonic branch the Mach number is after the shock is

Ty ρy Py P0y Mx My Tx ρx Px P0x 3.0000 0.47519 2.6790 3.8571 10.3333 0.32834

With Mach number of M = 0.47519 the maximum heat transfer requires information from Rayleigh ﬂow as following

M T T0 P P0 ρ∗ T∗ T0∗ P∗ P0∗ ρ 0.33138 0.47519 0.40469 2.0802 1.1857 0.22844

M T T0 P P0 ρ∗ T∗ T0∗ P∗ P0∗ ρ 0.47519 0.75086 0.65398 1.8235 1.1244 0.41176

It also must be noticed that stagnation temperature remains constant across shock wave. T Q 1 01 m˙ T0∗ 1 0.65398 subsonic = − subsonic = − = 1 Q T01 1 0.65398 ˙ 1 m super sonic − T0∗ supersonic −

It is not surprising since the the shock wave are found on the Rayleigh ﬂow. 200 CHAPTER 10. RAYLEIGH FLOW CHAPTER 11

Evacuating and Filling a Semi Rigid Chambers

Fanno model

In some ways the next two Chapters for relatively short tube contain materials is new to the tradi- © 1 tional compressible ﬂow text books . It Isothermal model was the undersigned experience, that for relatively long tube in traditional classes for with compress- Volume forced models ible ﬂow (sometimes referred to as gas External forces that control the tank volume Fanno model dynamics) don’t provide a demonstra- for relativly short tube

tion to applicability of the class mate-

j #¡1¢#£"¤ rial aside to aeronautical spectrum even such as turbomachinery. In this Chap- Isothermal model for relativly long tube ter a discussion on application of com- Volume is a function of pressure or rigid pressible flow to other fields like manu- (the volume can be also a function of inertia and etc) facturing is presented2. Semi rigid tank There is a significant impor- Fig. 11.1: The two different classifications of tance to the “pure” models such Isother- models that explain the filling or mal flow and Fanno flow which have im- evacuating of a single chamber mediate applicability. However, in many instances, the situations, in life, are far

1After completion of these Chapters, the undersigned discover two text books which to include some material related to this topic. These books are OCR, J. A., Fundamentals of Gas Dynamics, Interna- tional Textbook Co., Scranton, Pennsylvania, 1964. and “Compressible Fluid Flow,” 2nd Edition, by M. A. Saad, Prentice Hall, 1985. However, these books contained only limit discussions on the evacuation of chamber with attached nozzle. 2Even if the instructor feels that their students are convinced about the importance of the compressible, this example can further strength and enhance this conviction.

201 202 CHAPTER 11. EVACUATING AND FILLING A SEMI RIGID CHAMBERS more complicate. Combination of gas compressibility in the chamber and ﬂow out or through a tube post a special interest and these next two Chapters are dealing with these topics. In the ﬁrst Chapter models, were the chamber volume is controlled or a function of the pressure, are discussed. In the second Chapter, models, were the chamber’s volume is a function of external forces, are presented (see Figure (11.1)).

11.1 Governing Equations and Assumptions The process of filing or evacuating a semi flexible (semi rigid) chamber through a tube is very common in engineering. For example, most car today equipped with an airbag. For instance, the models in this Chapter are suitable for study of the filling the airbag or filling bicycle with air. The analysis is extended to include a semi rigid tank. The term semi rigid tank referred to a tank that the volume is either completely rigid or is a function of the chamber’s pressure. As it was shown in this book the most appropriate model for the flow in the tube for a relatively fast situation is Fanno Flow. The Isothermal model is more appropriate for cases where the tube is relatively long in–which a significant heat transfer occurs keeping the temperature almost constant. As it was shown in Chap- 4fL ter (9) the resistance, D , should be larger than 400. Yet Isothermal flow model is used as the limiting case.

fanno model fanno model reduced for relatively short tube for relatively short tube

connection 1 2 1 2

²§³©´oµ ¶w·#¸¯¹2º¯»

¥§¦©¨oª «w¬#¯®2°¯±

Isothermal model Isothermal model for a relatively long tube for a relatively long tube

The connection is through a narrow passage A schematic of a direct connection

Fig. 11.2: A schematic of two possible connections of the tube to a single chamber

The Rayleigh ﬂow model requires

¼½I¾K¿ ÀÂÁ Ã Ä

Ç È that a constant heat transfer supplied either Å(Æ by chemical reactions or otherwise. This author isn’t familiar with situations in which Control volume for the evacuating case Rayleigh ﬂow model is applicable. And

therefore, at this stage, no discussion is of-

ÉÊIËKÌ ÍÂÎ Ï Ð Ó fered here. Ñ(Ò Fanno flow model is the most appropriate in the case where the filling and evacuating is relatively fast. In case the Control volume for the filling case filling is relatively slow (long 4fL than the D Fig. 11.3: A schematic of the control vol- Isothermal flow is appropriate model. Yet umes used in this model as it was stated before, here Isothermal flow and Fanno flow are used as limiting or bounding cases for the real flow. Addition- 11.1. GOVERNING EQUATIONS AND ASSUMPTIONS 203 ally, the process in the chamber can be limited or bounded between two limits of Isentropic process or Isothermal process. In this analysis, in order to obtain the essence of the process, some simplified assumptions are made. The assumptions can be relaxed or removed and the model will be more general. Of course, the payment is by far more complex model that sometime clutter the physics. First, a model based on Fanno flow model is constructed. Second, model is studied in which the flow in the tube is isothermal. The flow in the tube in many cases is somewhere between the Fanno flow model to Isothermal flow model. This reality is an additional reason for the construction of two models in which they can be compared. Effects such as chemical reactions (or condensation/evaporation) are neglected. There are two suggested itself possibilities to the connection between the tube to the tank (see the Figure 11.2): one) direct two) through a reduction. The direct connection is when the tube is connect straight to tank like in a case where pipe is welded into the tank. The reduction is typical when a ball is filled trough an one–way valve (filling a baseball ball, also in manufacturing processes). The second possibility leads itself to an additional parameter that is independent of the 4fL resistance. The first kind connection tied the resistance, D , with the tube area. The simplest model for gas inside the chamber as a first approximation is the isotropic model. It is assumed that kinetic change in the chamber is negligible. Therefore, the pressure in the chamber is equal to the stagnation pressure, P P ≈ 0 (see Figure (11.4)). Thus, the stagnation pressure at the tube’s entrance is the same as the pressure in the chamber.

The mass in the chamber and

mass ﬂow out are expressed in

Ø Ø terms of the chamber variables ÔXÕÖÔ*× (see Figure 11.3. The mass in ÙÛÚ

the tank for perfect gas reads ÜÞÝ©ß 1 2 dm m˙ out = 0 (11.1) dt − Fig. 11.4: The pressure assumptions in the chamber And for perfect gas the mass at and tube entrance any given time is

P (t)V (t) m = (11.2) RT (t)

4fL The mass ﬂow out is a function of the resistance in tube, D and the pressure 4fL difference between the two sides of the tube m˙ out( D , P1/P2). The initial conditions in the chamber are T (0), P (0) and etc. If the mass occupied in the tube is 204 CHAPTER 11. EVACUATING AND FILLING A SEMI RIGID CHAMBERS neglected (only for ﬁlling process) the most general equation ideal gas (11.1) reads

m˙ out

m U d P V P zρ A c M}|( 4fL , 2 {) = 0 (11.3) dt RT 1 1 1 D P z }| { z }| 1 { When the plus sign is for ﬁlling process and the negative sign is for evacuating process.

11.2 General Model and Non-dimensioned It is convenient to non-dimensioned the properties in chamber by dividing them by their initial conditions. The dimensionless properties of chamber as T (t = t¯) T¯ = (11.4a) T (t = 0) V (t = t¯) V¯ = (11.4b) V (t = 0) P (t = t¯) P¯ = (11.4c) P (t = 0) t t¯= (11.4d) tc where tc is the characteristic time of the system deﬁned as followed V (0) tc = (11.5) AMmax kRT (0))

The physical meaning of characteristic timep, tc is the time that will take to evacuate the chamber if the gas in the chamber was in its initial state, the flow rate was at its maximum (choking flow), and the gas was incompressible in the chamber. Utilizing these definitions (11.4) and substituting into equation (11.3) yields

ρ c(t) ¯ ¯ ¯ P (0)V (0) d P V P1 P (0) ¯ ¯ ¯ ¯ ¯ A kRT1T (0)MmaxM(t) = 0 (11.6) tcRT (0) dt¯ T zRT1}|T (0){ qz }| { where the following deﬁnition for the reduced Mach number is added as M (t) M¯ = 1 (11.7) Mmax After some rearranging equation (11.6) obtains the form

d P¯V¯ tcAMmax kRT (0) P¯ M¯ 1 1 M¯ = 0 (11.8) dt¯ T¯ V (0) ¯ p T1 p 11.2. GENERAL MODEL AND NON-DIMENSIONED 205 and utilizing the deﬁnition of characteristic time, equation (11.5), and substituting into equation (11.8) yields d P¯V¯ P¯ M¯ 1 = 0 (11.9) dt¯ T¯ ¯ T1 Note that equation (11.9) can be modiﬁedpby introducing additional parameter 3 which referred to as external time, tmax . For cases, where the process time is important parameter equation (11.9) transformed to d P¯V¯ t P¯ M¯ max 1 = 0 (11.10) dt˜ T¯ t ¯ c T1 when P¯, V¯ , T¯, and M¯ are all are function of t˜pin this case. And where t˜= t/tmax. It is more convenient to deal with the stagnation pressure then the actual pressure at the entrance to the tube. Utilizing the equations developed in Chapter 4 between the stagnation condition, denoted without subscript, and condition in a tube ¯ denoted with subscript 1. The ratio of P1 is substituted by √T¯1

(k+1) − 2(k 1) P¯ P¯ k 1 − 1 = 1 + − M 2 (11.11) ¯ √ ¯ 2 T1 T It is convenient to denotep (k+1) − 2(k 1) k 1 − f[M] = 1 + − M 2 (11.12) 2 Note that f[M] is a function of the time. Utilizing the deﬁnitions (11.11) and substituting equation (11.12) into equation (11.9) to be transformed into d P¯V¯ P¯M¯ (t¯)f[M] = 0 (11.13) dt¯ T¯ √ ¯ T Equation (11.13) is a ﬁrst order nonlinear differential equation that can be solved for different initial conditions. At this stage, the author isn’t aware that there is a general solution for this equation4. Nevertheless, many numerical methods are available to solve this equation.

11.2.1 Isentropic Process The relationship between the pressure and the temperature in the chamber can be approximated as isotropic and therefore

k 1 − T (t) P (t) k k 1 T¯ = = = P¯ −k (11.14) T (0) P (0) 3This notation is used in many industrial processes where time of process referred to sometime as the maximum time. 4To those mathematically included, ﬁnd the general solution for this equation. 206 CHAPTER 11. EVACUATING AND FILLING A SEMI RIGID CHAMBERS The ratios can be expressed in term of the reduced pressure as followed:

P¯ P¯ 1 = = P¯ k (11.15) ¯ k 1 T P¯ −k and

P¯ k+1 = P¯ 2k (11.16) √T¯

So equation (11.13) is simpliﬁed into three different forms:

d 1 k+1 V¯ P¯ k P¯ 2k M¯ (t¯)f[M] = 0 (11.17a) dt¯

1 1 k dP¯ 1 dV¯ k+1 P¯ −k V¯ + P¯ k P¯ 2k M¯ (t¯)f[M] = 0 (11.17b) k dt¯ dt¯

dP¯ dV¯ 3k 1 V¯ + kP¯ kP¯ 2−k M¯ (t¯)f[M] = 0 (11.17c) dt¯ dt¯ Equation (11.17) is a general equation for evacuating or filling for isentropic process in the chamber. It should be point out that, in this stage, the model in the tube could be either Fanno flow or Isothermal flow. The situations where the chamber undergoes isentropic process but the flow in the tube is Isothermal are limited. Nevertheless, the application of this model provide some kind of a limit where to expect when some heat transfer occurs. Note the temperature in the tube entrance can be above or below the surrounding temperature. Simplified calculations of the entrance Mach number are described in the advance topics section.

11.2.2 Isothermal Process in The Chamber 11.2.3 A Note on the Entrance Mach number

4fL The value of Mach number, M1 is a function of the resistance, D and the ratio of pressure in the tank to the back pressure, PB /P1. The exit pressure, P2 is different from PB in some situations. As it was shown before, once the flow became 4fL choked the Mach number, M1 is only a function of the resistance, D . These statements are correct for both Fanno flow and the Isothermal flow models. The method outlined in Chapters 8 and 9 is appropriate for solving for entrance Mach number, M1. Two equations must be solved for the Mach numbers at the duct entrance and exit when the flow is in a chokeless condition. These equations are combinations of 11.3. RIGID TANK WITH NOZZLE 207 the momentum and energy equations in terms of the Mach numbers. The characteristic equations for Fanno flow (9.50), are

4fL = 4fLmax 4fLmax (11.18) D D 1 − D 2 h i h i and

k+1 k k 1 1 k k 1 2 − P k 1 − M 1 + − M2 2 = 1 + M 2 1 2 − 2 v k 1 2 (11.19) P0(t) 2 M2 u"1 + − M1 # u 2 t 4fL where D is deﬁned by equation (9.49). The solution of equations (11.18) and (11.19) for given 4fL and Pexit yields the D P0(t) entrance and exit Mach numbers. See advance topic about approximate solution 4fL for large resistance, D or small entrance Mach number, M1.

11.3 Rigid Tank with Nozzle he most simplest possible combination is discussed here before going trough the more complex cases A chamber is filled or evacuated by a nozzle. The gas in the chamber assumed to go an isentropic processes and flow is bounded in nozzle between isentropic flow and isothermal flow5. Here, it also will be assumed that the flow in the nozzle is either adiabatic or isothermal.

11.3.1 Adiabatic Isentropic Nozzle Attached The mass ﬂow out is given by either by Fliegner’s equation (4.47) or simply use cMρA∗ and equation (11.17) becomes

1 1 k dP¯ k+1 P¯ −k P¯ 2k (t¯)f[M] = 0 (11.20) k dt¯ It was utilized that V¯ = 1 and M¯ definition is simplified as M¯ = 1. It can be noticed that the characteristic time defined in equation (11.5) reduced into: V (0) tc = (11.21) A kRT (0))

Also it can be noticed that equation (11.12)p simpliﬁed into

(k+1) (k+1) −2(k 1) −2(k 1) k 1 − k + 1 − f[M] = 1 + − 12 = (11.22) 2 2

5This work is suggested by Donald Katze the point out that this issue appeared in Shapiro’s Book Vol 1, Chapter 4, p. 111 as a question 4.31. 208 CHAPTER 11. EVACUATING AND FILLING A SEMI RIGID CHAMBERS Equation (11.20) can be simpliﬁed as

1 1 k P 2−k dP f[m]dt¯= 0 (11.23) k Equation (11.23) can be integrated as

P¯ t 1 k P 2−k dP dt = 0 (11.24) Z1 Z0 The integration limits are obtained by simply using the deﬁnitions of reduced pressure, at P (t¯= 0) = 1 and P (t¯= t¯) = P¯. After the integration, equation (11.24) and rearrangement becomes

2k 1 k k 1 − P¯ = 1 − f[M]t¯ (11.25) 2

Example 11.1: A chamber is connected to a main line with pressure line with a diaphragm and nozzle. The initial pressure at the chamber is 1.5[Bar] and the volume is 1.0[m3]. Calculate time it requires that the pressure to reach 5[Bar] for two different nozzles throat area of 0.001, and 0.1 [m2] when diaphragm is erupted. Assumed the stagnation temperature at the main line is the ambient of 27[◦C].

SOLUTION The characteristic time is V V 1.0 tmax = = = = 0.028[sec] (11.26) A c A c 0.1√1.4 287 300 ∗ ∗ × × And for smaller area 1.0 tmax = = 2.8[sec] 0.001√1.4 287 300 × ×

P (t) 4.5 P¯ = = = 3.0 P (0) 1.5 The time is () 1 k k + 1 − t = t P¯ −k 1 (11.27) max − 2 h i Substituting values into equation (11.27) results

2.4 − 1 1.4 2.4 0.8 t = 0.028 3 −2.8 1 = 0.013[sec] (11.28) − 2 h i 11.4. RAPID EVACUATING OF A RIGID TANK 209

Filling/Evacuating The Chamber Under Upchucked Condition

The ﬂow in the nozzle can became upchucked and it can be analytically solved. Owczarek [1964] found an analytical solution which described here.

11.3.2 Isothermal Nozzle Attached

In this case the process in nozzle is assumed to isothermal but the process in the chamber is isentropic. The temperature in the nozzle is changing because the temperature in the chamber is changing. Yet, the differential temperature change in the chamber is slower than the temperature change in nozzle. For rigid volume, V¯ = 1 and for isothermal nozzle T¯ = 1 Thus, equation (11.13) is reduced into

dP¯ = f[M]P¯ = 0 (11.29) dt¯ Separating the variables and rearranging equation (11.29) converted into

¯ P dP¯ t¯ f[M] dt¯= 0 (11.30) P¯ Z1 Z0 Here, f[M] is expressed by equation (11.22). After the integration, equation (11.30) transformed into

(k+1) − 2(k 1) k + 1 − ln P¯ = t¯ 2 (k+1) k+1 −2(k 1) ¯ ( 2 ) − t P¯ = e" # (11.31)

11.4 Rapid evacuating of a rigid tank 11.4.1 With Fanno Flow

The relative Volume, V¯ (t) = 1, is constant and equal one for a completely rigid tank. In such case, the general equation (11.17) “shrinks” and doesn’t contain the relative volume term. A reasonable model for the tank is isentropic (can be replaced polytropic relationship) and Fanno flow are assumed for the flow in the tube. Thus, the specific governing equation is

dP¯ 3k 1 kM¯ f[M]P¯ 2−k = 0 (11.32) dt¯ − 210 CHAPTER 11. EVACUATING AND FILLING A SEMI RIGID CHAMBERS

For a choked ﬂow the entrance Mach number to the tube is at its maximum, Mmax and therefore M¯ = 1. The solution of equation (11.32) is obtained by noticing that M¯ is not a function of time and by variables separation results in t¯ P¯ P¯ dP¯ 1 1 3k ¯ ¯ −2k ¯ dt = 3k 1 = P dP (11.33) ¯ ¯ 2−k kM¯ f[M] Z0 Z1 kMf[M]P Z1 direct integration of equation (11.33) results in

2 1 k t¯= P¯ 2−k 1 (11.34) (k 1)M¯ f[M] − − h i It has to be realized that this is “reversed” function i.e. t¯ is a function of P and can be reversed for case. But for the chocked case it appears as

2k 1 k (k 1)M¯ f[M] − P¯ = 1 + − t¯ (11.35) 2 The function is drawn as shown here in Figure (11.5). The Figure (11.5) shows

1.0

P(t) V(t) = P (t) 0.8

V(t) = P (0)

0.6

0.4

0.2

0 0 0.2 0.4 0.6 0.8 1.0

t¯

Fig. 11.5: The reduced time as a function of the modiﬁed reduced pressure

that when the modified reduced pressure equal to one the reduced time is zero. The reduced time increases with decrease of the pressure in the tank. At certain point the flow becomes chokeless flow (unless the back pressure is The big struggle look for sug- complete vacuum). The transition point is denoted here as chT . Thus, equation gestion for better notation. (11.34) has to include the entrance Mach under the integration sign as P¯ 1 1 3k t¯ t¯ = P¯ −2k dP¯ (11.36) − chT kM¯ f[M] ZPchT 11.4. RAPID EVACUATING OF A RIGID TANK 211 For practical purposes if the flow is choked for more than 30% of the charecteristic time the choking equation can be used for the whole range, unless extra long time or extra low pressure is calculated/needed. Further, when the flow became chokeless the entrance Mach number does not change much from the choking condition. Again, for the special cases where the choked equation is not applicable the integration has to be separated into zones: choked and chokeless flow regions. And in the choke region the calculations can use the choking formula and numerical calculations for the rest.

Example 11.2: A chamber with volume of 0.1[m3] is ﬁlled with air at pressure of 10[Bar]. The chamber is connected with a rubber tube with f = 0.025, d = 0.01[m] and length of L = 5.0[m]

SOLUTION 4fL 4fL The ﬁrst parameter that calculated is D D = 5

11.4.2 Filling Process

The governing equation is

dP¯ 3k 1 kM¯ f[M]P¯ 2−k = 0 (11.37) dt¯ −

For a choked ﬂow the entrance Mach number to the tube is at its maximum, Mmax and therefore M¯ = 1. The solution of equation (11.37) is obtained by noticing that M¯ is not a function of time and by variable separation results in

t¯ P¯ P¯ dP¯ 1 1 3k ¯ ¯ −2k ¯ dt = 3k 1 = P dP (11.38) ¯ ¯ 2−k kM¯ f[M] Z0 Z1 kMf[M]P Z1 direct integration of equation (11.38) results in

2 1 k t¯= P¯ 2−k 1 (11.39) (k 1)M¯ f[M] − − h i It has to be realized that this is a reversed function. Nevertheless, with today computer this should not be a problem and easily can be drawn as shown here in Figure (11.5). The Figure shows that when the modified reduced pressure equal to one the reduced time is zero. The reduced time increases with decrease of the pressure in the tank. At some point the flow becomes chokeless flow (unless the back pressure is a complete vacuum). The transition point is denoted here as chT . Thus, equation 212 CHAPTER 11. EVACUATING AND FILLING A SEMI RIGID CHAMBERS

1 V(t) = P(t) V(t) = V(0)

0.8

0.6

0.4

0.2

0 0.2 0.4 0.6 0.8 1

àáPâäã3å¯æ(ç èéwêìë3í

ð ñòIóõô/ö"ñø÷ùòPúûô/ö¯ü(ý þ

ï ÿ

î or

ò¡

Fig. 11.6: The reduced time as a function of the modiﬁed reduced pressure

(11.39) has to include the entrance Mach under the integration sign as

P¯ 1 1 3k t¯ t¯ = P¯ −2k dP¯ (11.40) − chT kM¯ f[M] ZPchT

11.4.3 The Isothermal Process

For Isothermal process, the relative temperature, T¯ = 1. The combination of the isentropic tank and Isothermal flow in the tube is different from Fanno flow in that the chocking condition occurs at 1/√k. This model is reasonably appropriated when the chamber is insulated and not flat while the tube is relatively long and the process is relatively long. It has to be remembered that the chamber can undergo isothermal process. For the double isothermal (chamber and tube) the equation (11.6) reduced into

ρ c(0) P (0)V (0) d P¯V¯ P¯ P (0) 1 A kRT (0)M M¯ (t¯) = 0 ¯ max (11.41) tcRT (0) dt zR }|T (0){ pz }| { 11.4. RAPID EVACUATING OF A RIGID TANK 213 11.4.4 Simple Semi Rigid Chamber A simple relation of semi rigid chamber when the volume of the chamber is linearly related to the pressure as

V (t) = aP (t) (11.42)

where a is a constant that represent the physics. This situation occurs at least in small ranges for airbag balloon etc. The physical explanation when it occurs beyond the scope of this book. Nevertheless, a general solution is easily can be obtained similarly to rigid tank. Substituting equation (11.42) into yields

d 1+k k+1 P¯ k P¯ 2k M¯ f[M] = 0 (11.43) dt¯ − Carrying differentiation result in

1 + k 1 dP¯ k+1 P¯ k P¯ 2k M¯ f[M] = 0 (11.44) k dt¯ −

Similarly as before, the variables are separated as

t¯ P¯ k 1 k P¯ 2−k dP¯ dt = (11.45) 1 + k M¯ f[M] Z0 Z1 The equation (11.45) integrated to obtain the form

2 2k 3k 1 t¯= 1 P¯ 2−k (11.46) M¯ f[M](3k 1)(1 + k) − − h i The physical meaning that the pressure remains larger thorough evacuating process, as results in faster reduction of the gas from the chamber.

11.4.5 The “Simple” General Case The relationship between the pressure and the volume from the physical point of view must be monotonous. Further, the relation must be also positive, increase of the pressure results in increase of the volume (as results of Hook’s law. After all, in the known situations to this author pressure increase results in volume decrease (at least for ideal gas.). In this analysis and previous analysis the initial effect of the chamber container inertia is neglected. The analysis is based only on the mass conservation and if unsteady effects are required more terms (physical quantities) have taken into account. Further, it is assumed the ideal gas applied to the gas and this assumption isn’t relaxed here. 214 CHAPTER 11. EVACUATING AND FILLING A SEMI RIGID CHAMBERS Any continuous positive monotonic function can be expressed into a polynomial function. However, as ﬁrst approximation and simpliﬁed approach can be done by a single term with a different power as

V (t) = aP n (11.47)

When n can be any positive value including zero, 0. The physical meaning of n = 0 is that the tank is rigid. In reality the value of n lays between zero to one. When n is approaching to zero the chamber is approaches to a rigid tank and vis versa when the n 1 the chamber is ﬂexible like a balloon. → There isn’t a real critical value to n. Yet, it is convenient for engineers to further study the point where the relationship between the reduced time and the reduced pressure are linear6 Value of n above it will Convex and below it concave.

d 1+nk k k+1 P¯ k − P¯ 2k M¯ f[M] = 0 (11.48) dt¯ −

Notice that when n = 1 equation (11.49) reduced to equation (11.43). After carrying–out differentiation results

1 + nk k 1+nk 2k dP¯ k+1 − P¯ k− P¯ 2k M¯ f[M] = 0 (11.49) k dt¯ −

Again, similarly as before, variables are separated and integrated as follows

t¯ P¯ 1+2nk 5k 1 + nk k P¯ 2k− dP¯ dt = − (11.50) k M¯ f[M] Z0 Z1 Carrying–out the integration for the initial part if exit results in

2 2k 3k 2nk 1 t¯= 1 P¯ −2k − (11.51) M¯ f[M](3k 2nk 1)(1 + k) − − − h i The linear condition are obtain when

3k 2 3k 2nk 1 = 1 n = − (11.52) − − −→ 2k

That is just bellow 1 (n = 0.785714286) for k = 1.4.

6Some suggested this border point as infinite evocation to infinite time for evacuation etc. This undersigned is not aware situation where this indeed play important role. Therefore, it is waited to find such conditions before calling it as critical condition. 11.5. ADVANCE TOPICS 215 11.5 Advance Topics 4fL The term D is very large for small values of the entrance Mach number which requires keeping many digits in the calculation. For small values of the Mach numbers, equation (11.18) can be approximated as

2 2 4fL 1 Mexit Min D = 2− 2 (11.53) k Mexit Min and equation (11.19) as P M exit = in . (11.54) P0(t) Mexit

The solution of two equations (11.53) and (11.54) yields

2 1 Pexit P0(t) M = v − . (11.55) in u h4fL i u k D u t This solution should used only for Min < 0.00286; otherwise equations (11.18) and (11.19) must be solved numerically. The solution of equation (11.18) and (11.19) is described in “Pressure die casting: a model of vacuum pumping” Bar-Meir, G; Eckert, E R G; Goldstein, R. J. Journal of Manufacturing Science and Engineering (USA). Vol. 118, no. 2, pp. 259-265. May 1996. 216 CHAPTER 11. EVACUATING AND FILLING A SEMI RIGID CHAMBERS CHAPTER 12

Evacuating/Filing Chambers under External Volume Control

This chapter is the second on the section dealing with filling and evacuating chambers. Here the model deals with the case where the volume is controlled by external forces. This kind of model is applicable to many manufacturing processes such as die casting, extraction etc. In general the process of the displacing the gas (in many cases air) with a liquid is a very common process. For example, in die casting process liquid metal is injected to a cavity and after the cooling/solidification period a part is obtained in near the final shape. One can also view the exhaust systems of internal combustion engine in the same manner. In these processes, sometime is vital to obtain a proper evacuation of the gas (air) from the cavity.

12.1 General Model In this analysis, in order to obtain the essence of the process, some simpliﬁed assumptions are made. It simplest model of such process is when a piston is displacing the gas though a long tube. It assumed that no chemical reaction (or condensation/evaporation) occur in the piston or the tube 1. It is further assumed that the process is relatively fast. The last assumption is a appropriate assumption in process such as die casting. Two extreme possibilities again suggest themselves: rapid and slow processes. The two different connections, direct and through reduced area are com- again to add the dimensional analysis what is rapid and bined in this analysis. what is slow.

1such reaction are possible and expected to be part of process but the complicates the analysis and do not contribute to understand to the compressibility effects.

217 218CHAPTER 12. EVACUATING/FILING CHAMBERS UNDER EXTERNAL VOLUME CONTROL 12.1.1 Rapid Process

Clearly under the assumption of rapid process the heat transfer can be neglected and Fanno ﬂow can be assumed for the tube. The ﬁrst approximation isotropic process describe the process inside the cylinder (see Figure (12.1)).

Fanno model

Fig. 12.1: The control volume of the “Cylinder”

Before introducing the steps of the analysis, it is noteworthy to think about the process in qualitative terms. The replacing incompressible liquid enter in the same amount as replaced incompressible liquid. But in a compressible substance the situation can be totally different, it is possible to obtain a situation where that most of the liquid entered the chamber and yet most of the replaced gas can be still be in the chamber. Obtaining conditions where the volume of displacing liquid is equal to the displaced liquid are called the critical conditions. These critical conditions are very significant that they provide guidelines for the design of processes. Obviously, the best ventilation is achieved with a large tube or area. In manufacture processes to minimize cost and the secondary machining such as trimming and other issues the exit area or tube has to be narrow as possible. In the exhaust system cost of large exhaust valve increase with the size and in addition reduces the strength with the size of valve2. For these reasons the optimum size is desired. The conflicting requirements suggest an optimum area, which is also indicated by experimental studies and utilized by practiced engineers. The purpose of this analysis to yields a formula for critical/optimum vent area in a simple form is one of the objectives of this section. The second objective is to provide a tool to “combine” the actual tube with the resistance in the tube, thus, eliminating the need for calculations of the gas flow in the tube to minimize the numerical calculations. A linear function is the simplest model that decibels changes the volume. In reality, in some situations like die casting this description is appropriate. Nev- ertheless, this model can be extended numerical in cases where more complex function is applied.

t V (t) = V (0) 1 (12.1) − t max 2After certain sizes, the possibility of crack increases. 12.1. GENERAL MODEL 219 Equation (12.1) can be non–dimensionlassed as V¯ (t¯) = 1 t¯ (12.2) −

The governing equation (11.10) that was developed in the previous Chap- ter (11) obtained the form as

1 1 V¯ dP¯ dV¯ t M¯ f(M) k+1 P¯ k + + max P¯ 2k = 0 (12.3) k P¯ dt dt¯ t c where t¯ = t/tmax. Notice that in this case that there are two different characteristic times: the “characteristic” time, tc and the “maximum” time, tmax. The first characteristic time, tc is associated with the ratio of the volume and the tube characteristics (see equation (11.5)). The second characteristic time, tmax is associated with the imposed time on the system (in this case the elapsed time of the piston stroke). Equation (12.3) is an nonlinear first order differential equation and can be rearranged as follows dP¯ dt¯ = ; P¯(0) = 1. (12.4) t k 1 ¯ k 1 max M¯ f[M]P¯ 2−k P¯ 1 t − tc − Equation (12.4) is can be solved only when the flow is chocked In which case f[m] isn’t function of the time. The solution of equation (12.4)) can be obtained by transforming and by k 1 2k − k 1 introducing a new variable ξ = P¯ 2k and therefore P¯ = [ξ] − . The reduced Pres- 2k 2k 1 ¯ ( k 1 )− sure derivative, dP = k 1 [ξ] − dξ Utilizing this definition and there implication reduce equation (12.4)−

2k ( k 1 ) 1 2 [ξ] − − dξ dt¯ 2k = (12.5) k 1 1 t¯ (k 1) (1 Bξ) [ξ] − − − − where B = tmax M¯ f[M] And equation (12.5) can be further simpliﬁed as tc 2dξ dt¯ = (12.6) (k 1) (1 Bξ) ξ 1 t¯ − − − Equation (12.6) can be integrated to obtain 2 1 Bξ ln − = ln t¯ (12.7) (k 1)B ξ −

− or in a different form 2 (1 k)B 1 Bξ − − = t¯ (12.8) ξ

220CHAPTER 12. EVACUATING/FILING CHAMBERS UNDER EXTERNAL VOLUME CONTROL Now substituting to the “preferred” variable

2 1 k 1 t tmax − (1 k) max M¯ f[M] 1 M¯ f[M]P¯ 2k − tc tc ¯ − k 1 = t (12.9) ¯ − " P 2k # P¯

The analytical solution is applicable only in the case which the ﬂow is choked thorough all the process. The solution is applicable to indirect connection. This happen when vacuum is applied outside the tube (a technique used in die casting and injection molding to improve quality by reducing porosity.). In case when the ﬂow chokeless a numerical integration needed to be performed. In the literature, to create a direct function equation (12.4) is transformed into

t k 1 k 1 max M¯ f[M]P¯ 2−k dP¯ tc = − (12.10) dt¯ 1 t¯ − with the initial condition of

P (0) = 1 (12.11)

The analytical solution also can be approximated by a simpler equation as

tmax P¯ = [1 t] tc (12.12) − The results for numerical evaluation in the case when cylinder is initially at an atmospheric pressure and outside tube is also at atmospheric pressure are presented in Figure (12.2). In this case only some part of the ﬂow is choked (the later part). The results of a choked case are presented in Figure (12.3) in which outside tube condition is in vacuum. These Figures (12.2) and 12.3 demonstrate the importance of the ratio of tmax . When tmax > 1 the pressure increases signiﬁcantly tc tc and verse versa. Thus, the question remains how the time ratio can be transfered to parameters that can the engineer can design in the system. Denoting the area that creates the ratio tmax = 1 as the critical area, A tc c provides the needed tool. Thus the exit area, A can be expressed as A A = Ac (12.13) Ac

The actual times ratio tmax can be expressed as tc @A

1 tmax tmax tmax = (12.14) tc tc tc @A @A z }| @A{c

12.1. GENERAL MODEL 221 According to equation (11.5) t is inversely proportional to area, t 1/A. Thus, c c ∝ equation (12.14) the tmax is canceled and reduced into t A max = (12.15) t A c @A c

Parameters influencing the process are the area ratio, A , and the friction Ac 4fL parameter, D . From other detailed calculations the author thesis (later to be published on this www.potto.org). it was found that the influence of the parameter 4fL D on the pressure development in the cylinder is quite small. The influence is small on the residual air mass in the cylinder but larger on the Mach number, M . The effects of the area ratio, A , are studied here since it is the dominant exit Ac parameter. It is important to point out the significance of the tmax . This parameter rep- tc resents the ratio between the filling time and the evacuating time, the time which would be required to evacuate the cylinder for constant mass flow rate at the maximum Mach number when the gas temperature and pressure remain in their initial values. This parameter also represents the dimensionless area, A , according to Ac the following equation Figure (12.4) describes the pressure as a function of the dimensionless time for various values of A . The line that represents A = 1 is almost straight. Ac Ac For large values of A the pressure increases the volume flow rate of the air until Ac a quasi steady state is reached. This quasi steady state is achieved when the volumetric air flow rate out is equal to the volume pushed by the piston. The pressure and the mass flow rate are maintained constant after this state is reached. The pressure in this quasi steady state is a function of A . For small values of A there Ac Ac is no steady state stage. When A is greater than one the pressure is concave up- Ac ward and when A is less than one the pressure is concave downward as shown Ac in Figures (12.4), which was obtained by an integration of equation (12.9).

12.1.2 Examples Example 12.1: Calculate the minimum required vent area for die casting process when the die vol- 3 4fL ume is 0.001[m ] and D = 20. The required solidiﬁcation time, tmax = 0.03[sec].

SOLUTION

12.1.3 Direct Connection In the above analysis is applicable to indirect connection. It should be noted that critical area, Ac, is not function of the time. The direct connection posts more mathematical difﬁculty because the critical area is not constant and time dependent. 222CHAPTER 12. EVACUATING/FILING CHAMBERS UNDER EXTERNAL VOLUME CONTROL To continue

12.2 Summary The analysis indicates there is a critical vent area below which the ventilation is poor and above which the resistance to air flow is minimal. This critical area depends on the geometry and the filling time. The critical area also provides a mean to “combine” the actual vent area with the vent resistance for numerical simulations of the cavity filling, taking into account the compressibility of the gas flow. 12.2. SUMMARY 223

5.0 Dimensionless ∗

4.6 Area, A/Ac ∇

∗

4.2 0.0 ∗ ∇

0.2 ∗

3.8 0.5 ∇ ∇ ∗ 1.0 3.4 1.2 ∇

∗ 2.0 o ∇ P (t) ∗ 3.0 5.0 ∇ P (0)

∗ ∇

4fL ∇ 2.6 = 100.0 ∗ D ∇ ∗ ∇ ∗ ∇ 2.2 ∇ ∗ ∇

∗ ∇

∇

∗

∇

∗

∇

1.8

∇

∗

∇

∗

∇

∗

∇

∗

∇

∗ ∇ 1.4 ∇ ∗ ∇

∇ ∗

∗ ∇

∗ ∇ ∇ ∇∗ ∇ o o o o o o o o o o o o o o o o o o o o o o o o o ∇∗ ∇∗ o o o o o o o o o 1.0 ∇o∗ ∇o∗ o o 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

Figure a

5.0 Dimensionless ∗ 4.6 Area, A/A

c ∇ ∗

0.0 ∗

4.2 ∇

0.2 ∗

3.8 0.5 ∇ ∇

1.0 ∗

3.4 1.2 ∇

∗ 2.0 o ∇ P (t) ∗ 3.0 5.0 ∇ P (0)

∗ ∇ 4fL = ∗ 5.0 ∇ 2.6 D ∗ ∇

∇ ∗ 2.2 ∇

∗ ∇

∇ ∗

∇

∗

1.8

∇

∗

∇

∗

∇

∗

∇

∗ ∇

∇

∗ 1.4 ∇ ∗ ∇

∇

∗ ∇

∗ ∇ ∇ ∇∗ ∇ ∇∗ ∇∗ o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o 1.0 ∇o∗ ∇o∗ o o 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 Dimensionless Time, t, or, Cylinder Volume Fraction Figure b

Fig. 12.2: The pressure ratio as a function of the dimensionless time for chokeless condition 224CHAPTER 12. EVACUATING/FILING CHAMBERS UNDER EXTERNAL VOLUME CONTROL

2.2

∗ . .. 2 .. ∗ .. .. ∗ .. .. ∗ .. 1.8 .. ∗ .. .. ∗ .. .. ∗

. A . 1.6 .. ∗ = 0.0 .. ∗ A .. c .. ∗ 0.1 ...... ∗ .. .. ∗ 1.4 .. ∗ 0.5 ∗ .. ∗ ... ∗ ... ∗ 1.0 .. ∗ .. ∗ 1.2 .. ∗ 1.5 × .. ∗ ... ∗ ∗ . .. ∗ ∗ 4.0 ... ∗ P(t) ...∗ ∗ . 1 ∗×. ∗ × × P(0) × × × × × × × × × 0.8 × ×

× × × × × × × × × × 0.6 × × × ×

× × × × ×

× 0.4 ×

× ×

0.2

0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 DIMENSIONLESS TIME, t, or, CYLINDER VOLUME FRACTION

Fig. 12.3: The pressure ratio as a function of the dimensionless time for choked condition

Fig. 12.4: The pressure ratio as a function of the dimensionless time CHAPTER 13

Topics in Unsteady one Dimensional gas dynamics

characteristic equations and the method of characteristics, The unsteady effect of the compressible ﬂow are more complicated than in incompressible ﬂow and a special techniques are need. The technique of characteristics is introduced. First the physical case which is analyzed is described. A tube with a membrane as shown in Figure

225 226CHAPTER 13. TOPICS IN UNSTEADY ONE DIMENSIONAL GAS DYNAMICS CHAPTER 14

Oblique-Shock

14.1 Preface to Oblique Shock In Chapter (5), discussion on a normal

shock was presented. A normal shock is a special type of shock wave. The other type of shock wave is the oblique shock. In the literature oblique shock,

normal shock, and Prandtl–Meyer function are presented as three separate = 0 and different issues. However, one can view all these cases as three different Fig. 14.1: A view of a straight normal shock as regions of a flow over a plate with a de- a limited case for oblique shock flection section. Clearly, variation of the deflection angle from a zero (δ = 0) to a positive value results in oblique shock. Further changing the deflection angle to a negative value results in expansion waves. The common representation is done by not showing the boundaries of these models. However, this section attempts to show the boundaries and the limits or connections of these models1.

1In this chapter, even the whole book, a very limited discussion about reflection shocks and collisions of weak shock, Von Neumann paradox, triple shock intersection, etc are presented. The author believes that these issues are not relevant to most engineering students and practices. Furthermore, these issues should not be introduced in introductory textbook of compressible flow. Those who would like to obtain more information, should refer to J.B. Keller, “Rays, waves and asymptotics,” Bull. Am. Math. Soc. 84, 727 (1978), and E.G. Tabak and R.R. Rosales, “Focusing of weak shock waves and the Von Neuman paradox of oblique shock reflection,” Phys. Fluids 6, 1874 (1994).

227 228 CHAPTER 14. OBLIQUE-SHOCK 14.2 Introduction 14.2.1 Introduction to Oblique Shock A normal shock occurs when there is a disturbance downstream which imposes a boundary condition on the flow in which the fluid/gas can react only by a sharp change in the flow direction. As it may be recalled, normal shock occurs when a wall is straight/flat (δ = 0) as shown in Figure (14.1) which occurs when somewhere downstream a disturbance2 appears. When the deflection angle is increased, the gas flow must match the boundary conditions. This matching can occur only when there is a discontinuity in the flow field. Thus, the direction of the flow is changed by a shock wave with an angle to the flow. This shock is commonly referred to as the oblique shock. Alternatively, as discussed in Chapter (1)3 the flow behaves as it does in a hyperbolic field. In such a case, the flow field is governed by a hyperbolic equation which deals with the case when information (like boundary conditions) reaches from downstream only if they are within the range of influence. For information such as the disturbance (boundary condition) to reach deep into the flow from the side requires time. During this time, the flow moves downstream and creates an angle.

14.2.2 Introduction to Prandtl–Meyer Function

Decreasing the deﬂection angle re- 0◦ Prandtl sults in the same effects as before. ν∞(k) Meyer θmax(k) Function Oblique The boundary conditions must match Shock the geometry. Yet, for a negative de-

flection angle (in this section’s nota- No Shock tion), the flow must be continuous. zone The analysis shows that the flow ve- Fig. 14.2: The regions where oblique shock or locity must increase to achieve this re- Prandtl–Meyer function exist. Notice quirement. This velocity increase is that both have a maximum point and referred to as the expansion wave. As a “no solution” zone, which is around it will be shown in the next chapter, zero. However, Prandtl-Meyer func- as opposed to oblique shock analy- tion approaches closer to a zero de- sis, the increase in the upstream Mach flection angle. number determines the downstream Mach number and the “negative” deflection angle. It has to be pointed out that both the oblique shock and the Prandtl–Meyer function have a maximum point for M . However, the maximum point for the 1 → ∞ Prandtl–Meyer function is much larger than the oblique shock by a factor of more than 2. What accounts for the larger maximum point is the effective turning (less entropy production) which will be explained in the next chapter (see Figure (14.2)).

2Zero velocity, pressure boundary conditions, and different inclination angle, are examples of forces that create shock. The zero velocity can be found in a jet ﬂowing into a still medium of gas. 3This section is under construction and does not appear in the book yet. 14.3. OBLIQUE SHOCK 229 14.2.3 Introduction to Zero Inclination

What happens when the inclination angle is zero? Which model is correct to use? Can these two conﬂicting models, the oblique shock and the Prandtl–Meyer function, co-exist? Or perhaps a different model better describes the physics. In some books and in the famous NACA report 1135 it was assumed that Mach wave and oblique shock co–occur in the same zone. Previously (see Chapter 5), it was assumed that normal shock occurs at the same time. In this chapter, the stability issue will be examined in greater detail.

14.3 Oblique Shock

The shock occurs in real- θ − δ

ity in situations where the $ %'&

( ) * + +

shock has three–dimensional θ

!#"

,-/. 0 -

effects. The three–dimensional 1#243 effects of the shock make it appear as a curved plane.

However, for a chosen arbitrary accuracy it requires a Comparsion Line specific small area, a one– dimensional shock can be considered. In such a case, Fig. 14.3: A typical oblique shock schematic the change of the orientation makes the shock considerations two–dimensional. Alternately, using an infinite (or a two–dimensional) object produces a two–dimensional shock. The two– dimensional effects occur when the flow is affected from the “side,” i.e., a change in the flow direction4. To match the boundary conditions, the flow turns after the shock to be parallel to the inclination angle. Figure (14.3) exhibits the schematic of the oblique shock. The deflection angle, δ, is the direction of the flow after the shock (parallel to the wall). The normal shock analysis dictates that after the shock, the flow is always subsonic. The total flow after the oblique shock can also be supersonic, which depends on the boundary layer. Only the oblique shock’s normal component undergoes the “shock.” The tangent component does not change because it does not “move” across the shock line. Hence, the mass balance reads

ρ1U1n = ρ2U2n (14.1)

4The author begs for forgiveness from those who view this description as offensive (There was an unpleasant email to the author accusing him of revolt against the holy of the holies.). If you do not like this description, please just ignore it. You can use the traditional explanation, you do not need the author’s permission. 230 CHAPTER 14. OBLIQUE-SHOCK The momentum equation reads

2 2 P1 + ρ1U1n = P2 + ρ2U2n (14.2)

The momentum equation in the tangential direction yields

U1t = U2t (14.3)

The energy balance reads

U 2 U 2 C T + 1n = C T + 2n (14.4) p 1 2 p 2 2 Equations (14.1), (14.2), and (14.4) are the same as the equations for normal shock with the exception that the total velocity is replaced by the perpendicular components. Yet the new relationship between the upstream Mach number, the deﬂection angle, δ, and the Mach angle, θ has to be solved. From the geometry it can be observed that U tan θ = 1n (14.5) U1t

and

U2 tan(θ δ) = n (14.6) − U2t

Unlike in the normal shock, here there are three possible pairs5 of solutions to these equations. The ﬁrst is referred to as the weak shock; the second is the strong shock; and the third is an impossible solution (thermodynamically)6. Experiments and experience have shown that the common solution is the weak shock, in which the shock turns to a lesser extent7. tan θ U = 1n (14.7) tan(θ δ) U − 2n The above velocity–geometry equations can also be expressed in term of Mach number, as M sin θ = 1n (14.8) M1

5This issue is due to R. Menikoff, who raised the solution completeness issue. 6The solution requires solving the entropy conservation equation. The author is not aware of “simple” proof and a call to ﬁnd a simple proof is needed. 7Actually this term is used from historical reasons. The lesser extent angle is the unstable angle and the weak angle is the middle solution. But because the literature referred to only two roots, the term lesser extent is used. 14.3. OBLIQUE SHOCK 231

M sin(θ δ) = 2n (14.9) − M2

M cos θ = 1t (14.10) M1

M cos(θ δ) = 2t (14.11) − M2

The total energy across an oblique shock wave is constant, and it follows that the total speed of sound is constant across the (oblique) shock. It should be noted that although, U = U the Mach number is M = M because the 1t 2t 1t 6 2t temperatures on both sides of the shock are different, T = T . 1 6 2 As opposed to the normal shock, here angles (the second dimension) have to be determined. The solution from this set of four equations, (14.8) through (14.11), is a function of four unknowns of M1, M2, θ, and δ. Rearranging this set utilizing geometrical identities such as sin α = 2 sin α cos α results in M 2 sin2 θ 1 tan δ = 2 cot θ 1 − (14.12) M 2 (k + cos 2θ) + 2 1 The relationship between the properties can be determined by substituting M1 sin θ for of M1 into the normal shock relationship, which results in P 2kM 2 sin2 θ (k 1) 2 = 1 − − (14.13) P1 k + 1 The density and normal velocity ratio can be determined by the following equation 2 2 ρ2 U1n (k + 1)M1 sin θ = = 2 2 (14.14) ρ1 U2n (k 1)M sin θ + 2 − 1 The temperature ratio is expressed as 2 2 2 T2 2kM1 sin θ (k 1) (k 1)M1 + 2 = − − 2 − (14.15) T1 (k + 1) M 1 Prandtl’s relation for oblique shock is k 1 U U = c2 − U 2 (14.16) n1 n2 − k + 1 t The Rankine–Hugoniot relations are the same as the relationship for the normal shock P P P P 2 − 1 = k 2 − 1 (14.17) ρ ρ ρ ρ 2 − 1 2 − 1 232 CHAPTER 14. OBLIQUE-SHOCK 14.4 Solution of Mach Angle Oblique shock, if orientated to a coordinate perpendicular and parallel shock plane is like a normal shock. Thus, the relationship between the properties can be determined by using the normal components or by utilizing the normal shock table developed earlier. One has to be careful to use the normal components of the Mach numbers. The stagnation temperature contains the total velocity. Again, the normal shock is a one–dimensional problem, thus, only one parameter is required (to solve the problem). Oblique shock is a two–dimensional problem and two properties must be provided so a solution can be found. Prob- ably, the most useful properties are upstream Mach number, M1 and the deﬂec- tion angle, which create a somewhat complicated mathematical procedure, and this will be discussed later. Other combinations of properties provide a relatively simple mathematical treatment, and the solutions of selected pairs and selected relationships will be presented.

14.4.1 Upstream Mach Number, M1, and Deﬂection Angle, δ

Again, this set of parameters is, perhaps, the most common and natural to examine. Thompson (1950) has shown that the relationship of the shock angle is obtained from the following cubic equation:

3 2 x + a1x + a2x + a3 = 0 (14.18)

where

x = sin2 θ (14.19)

and

2 M1 + 2 2 a1 = 2 k sin δ (14.20) − M1 − 2 2 2M1 + 1 (k + 1) k 1 2 a2 = + + − sin δ (14.21) − M 4 4 M 2 1 1 cos2 δ a3 = 4 (14.22) − M1

Equation (14.18) requires that x has to be a real and positive number to obtain a real deﬂection angle8. Clearly, sin θ must be positive, and the negative sign refers to the mirror image of the solution. Thus, the negative root of sin θ must be disregarded

−1 8This point was pointed out by R. Menikoff. He also suggested that θ is bounded by sin 1/M1 and 1. 14.4. SOLUTION OF MACH ANGLE 233 The solution of a cubic equation such as (14.18) provides three roots9. These roots can be expressed as 1 x = a + (S + T ) (14.23) 1 −3 1

1 1 1 x = a (S + T ) + i√3(S T ) (14.24) 2 −3 1 − 2 2 − and 1 1 1 x = a (S + T ) i√3(S T ) (14.25) 3 −3 1 − 2 − 2 − Where

3 S = R + √D, (14.26) q

3 T = R √D (14.27) − q and where the deﬁnition of the D is

D = Q3 + R2 (14.28) and where the deﬁnitions of Q and R are

3a a 2 Q = 2 − 1 (14.29) 9 and 9a a 27a 2a 3 R = 1 2 − 3 − 1 (14.30) 54 Only three roots can exist for the Mach angle, θ. From a mathematical point of view, if D > 0, one root is real and two roots are complex. For the case D = 0, all the roots are real and at least two are identical. In the last case where D < 0, all the roots are real and unequal. The physical meaning of the above analysis demonstrates that in the range where D > 0 no solution can exist because no imaginary solution can exist10. D > 0 occurs when no shock angle can be found, so that the shock normal component is reduced to subsonic and yet parallel to the inclination angle.

9The highest power of the equation (only with integer numbers) is the number of the roots. For example, in a quadratic equation there are two roots. 10A call for suggestions, to explain about complex numbers and imaginary numbers should be included. Maybe insert an example where imaginary solution results in no physical solution. 234 CHAPTER 14. OBLIQUE-SHOCK Furthermore, only in some cases when D = 0 does the solution have a physical meaning. Hence, the solution in the case of D = 0 has to be examined in the light of other issues to determine the validity of the solution.

When D < 0, the three unique roots are reduced to two roots at least for the steady state because thermodynamics dictates11 that. Physically, it can be shown that the ﬁrst solution(14.23), referred sometimes as a thermodynamically unstable root, which is also related to a decrease in entropy, is “unrealistic.” Therefore, the ﬁrst solution does not occur in reality, at least, in steady–state situations. This root has only a mathematical meaning for steady–state analysis12.

These two roots represent two different situations. First, for the second root, the shock wave keeps the flow almost all the time as a supersonic flow and it is referred to as the weak solution (there is a small section that the flow is subsonic). Second, the third root always turns the flow into subsonic and it is referred to as the strong solution. It should be noted that this case is where entropy increases in the largest amount.

In summary, if a hand moves the shock angle starting from the deflection angle and reaching the first angle that satisfies the boundary condition, this situation is unstable and the shock angle will jump to the second angle (root). If an additional “push” is given, for example, by additional boundary conditions, the shock angle will jump to the third root13. These two angles of the strong and weak shock are stable for a two–dimensional wedge (see the appendix of this chapter for a limited discussion on the stability14).

11This situation is somewhat similar to a cubical body rotation. The cubical body has three symmetrical axes which the body can rotate around. However, the body will freely rotate only around two axes with small and large moments of inertia. The body rotation is unstable around the middle axes. The reader can simply try it. 12There is no experimental or analytical evidence, that the author has found, showing that it is totally impossible. The “unstable” terms can be thermodynamcily stable in unsteady case. Though, those who are dealing with rapid transient situations should be aware that this angle of oblique shock can exist. There is no theoretical evidence that showing that in strong unsteady state this angle is unstable. The shock will initially for a very brief time transient in it and will jump from this angle to the thermodynamically stable angles. 13See the hist/rical discussion on the stability. There are those who view this question not as a stability equation but rather as under what conditions a strong or a weak shock will prevail. 14This material is extra and not recommended for standard undergraduate students. 14.4. SOLUTION OF MACH ANGLE 235 14.4.2 When No Oblique Shock Exist or When D > 0

Large deﬂection angle for given, M1

The first range is when the deflection angle reaches above the maximum point. For a given upstream Mach number, M1, a change in the inclination angle requires a larger energy to change the flow direction. Once, the inclination angle reaches the “maximum potential energy,” a change in the flow direction is no longer possible. In the alternative view, the fluid “sees” the disturbance (in this case, Fig. 14.4: Flow around spherically the wedge) in front of it and hence the normal ◦ blunted 30 cone-cylinder shock occurs. Only when the fluid is away from with Mach number 2.0. It the object (smaller angle) liquid “sees” the ob- can be noticed that the ject in a different inclination angle. This differ- normal shock, the strong ent inclination angle is sometimes referred to shock, and the weak shock as an imaginary angle. coexist.

The simple procedure For example, in Figure (14.4) and (14.5), the imaginary angle is shown. The ﬂow is far away from the object and does not “see’ the object. For example, for, M1 the maximum deﬂection angle is calculated when 3 2 −→ ∞ D = Q + R = 0. This can be done by evaluating the terms a1, a2, and a3 for M = . 1 ∞ a = 1 k sin2 δ 1 − − (k + 1)2 sin2 δ a = 2 4 a3 = 0

With these values the coefﬁcients R and Q are

2 2 2 (k+1) sin δ 2 2 9( )(1 + k sin δ) 4 (2)( )(1 + k sin δ) R = − − − 54 and

(1 + k sin2 δ)2 Q = 9 Solving equation (14.28) after substituting these values of Q and R provides series of roots from which only one root is possible. This root, in the case k = 1.4, is just above δ π (note that the maximum is also a function of the heat ratio, k). max ∼ 4 236 CHAPTER 14. OBLIQUE-SHOCK

The fluid doesn’t ’’see’ the object

M∞ The fluid ‘‘sees’’ the object with "imaginary" inclanation }angle } Intermediate zone }

The fluid "sees" the object infront

Fig. 14.5: The view of a large inclination angle from different points in the ﬂuid ﬁeld.

The Procedure for Calculating The Maximum Deﬂection Point

The maximum is obtained when D = 0. When the right terms deﬁned in (14.20)-(14.21), (14.29), and (14.30) are substituted into this equation and utilizing the trigonometrical sin2 δ + cos2 δ = 1 and other trigonometrical identities results in Maximum Deﬂection Mach Number’s equation in which is

M 2 (k + 1) (M 2 + 1) = 2(kM 4 + 2M 2 1) (14.31) 1 1n 1n 1n − This equation and its twin equation can be obtained by an alternative procedure proposed by someone15 who suggested another way to approach this issue. It can be noticed that in equation (14.12), the deﬂection angle is a function of the Mach angle and the upstream Mach number, M1. Thus, one can conclude that the maximum Mach angle is only a function of the upstream Much number, M1. This can be shown mathematically by the argument that differentiating equation (14.12) and equating the results to zero creates relationship between the Mach number, M1 and the maximum Mach angle, θ. Since in that equation there appears only the heat ratio k, and Mach number, M1, θmax is a function of only these parameters.

15At ﬁrst, it was seen as C. J.Chapman, English mathematician to be the creator but later an earlier version by several months was proposed by Bernard Grossman. At this stage it is not clear who was the ﬁrst to propose it. 14.4. SOLUTION OF MACH ANGLE 237 The differentiation of the equation (14.12) yields

kM 4 sin4 θ + 2 (k+1) M 2 M 2 sin2 θ 1 + (k+1) M 2 d tan δ 1 − 2 1 1 − 2 1 = 2 2 (14.32) dθ kM 4 sin4θ (k 1) +(k+1) M1 M 2 sin2 θ 1 1 − − 4 1 − h i Because tan is a monotonous function, the maximum appears when θ has its maximum. The numerator of equation (14.32) is zero at different values of the denominator. Thus, it is sufﬁcient to equate the numerator to zero to obtain the maximum. The nominator produces a quadratic equation for sin2 θ and only the positive value for sin2 θ is applied here. Thus, the sin2 θ is

k+1 2 k 1 2 k+1 4 1 + 4 M1 + (k + 1) 1 + −2 M1 + 2 M1 2 − sin θmax = r h 2 i (14.33) kM1 Equation (14.33) should be referred to as the maximum’s equation. It should be noted that both the Maximum Mach Deflection equation and the maximum’s equation lead to the same conclusion that the maximum M1n is only a function of upstream the Mach number and the heat ratio k. It can be noticed that the Maximum 2 Deflection Mach Number’s equation is also a quadratic equation for M1n . Once M1n is found, then the Mach angle can be easily calculated by equation (14.8). To compare these two equations the simple case of Maximum for an infinite Mach number is examined. It must be pointed out that similar procedures can also be proposed (even though it does not appear in the literature). Instead, taking the derivative with respect to θ, a derivative can be taken with respect to M1. Thus, d tan δ = 0 (14.34) dM1 and then solving equation (14.34) provides a solution for Mmax. A simplified case of the Maximum Deflection Mach Number’s equation for large Mach number becomes

k + 1 M1n = M1 for M1 >> 1 (14.35) r 2k k+1 Hence, for large Mach numbers, the Mach angle is sin θ = 2k , which makes θ = 1.18 or θ = 67.79◦. q With the value of θ utilizing equation (14.12), the maximum deﬂection angle can be computed. Note that this procedure does not require an approximation of M1n to be made. The general solution of equation (14.31) is

(k + 1)M 2 + 1 + (M 2 M 2(k + 1)2 + 8(k2 1) + 16(1 + k) 1 1 1 − M1n = r q 2√k (14.36) 238 CHAPTER 14. OBLIQUE-SHOCK Note that Maximum Deﬂection Mach Number’s equation can be extended to deal with more complicated equations of state (aside from the perfect gas model). This typical example is for those who like mathematics.

Example 14.1: Derive the perturbation of Maximum Deﬂection Mach Number’s equation for the case of a very small upstream Mach number number of the form M1 = 1 + . Hint, Start with equation (14.31) and neglect all the terms that are relatively small.

SOLUTION under construction

The case of D 0 or 0 δ ≥ ≥ The second range in which D > 0 is when δ < 0. Thus, ﬁrst the transition line in which D = 0 has to be determined. This can be achieved by the standard mathematical procedure of equating D = 0. The analysis shows regardless of the value of the upstream Mach number D = 0 when δ = 0. This can be partially demonstrated by evaluating the terms a1, a2, and a3 for the speciﬁc value of M1 as following

2 M1 + 2 a1 = 2 M1 2 2M1 + 1 a2 = 4 − M1 1 a3 = 4 (14.37) −M1

With values presented in equations (14.37) for R and Q becoming

2 2 2 2 M1 +2 2M1 +1 1 M1 +2 9 M 2 M 4 27 M− 4 2 M 2 R = 1 1 − 1 − 1 54 2 2 2 2 2 2 9 M1 + 2 2M1 + 1 + 27M1 2M1 M1 + 2 = 6 − (14.38) 54M1

and

2 2 3 2M1 +1 M1 +2 3 M 4 M 2 Q = 1 − 1 (14.39) 9 14.4. SOLUTION OF MACH ANGLE 239 Substituting the values of Q and R equations (14.38) (14.39) into equation (14.28) provides the equation to be solved for δ.

3 2 2 3 2M1 +1 M1 +2 3 M 4 M 2 1 − 1 +  9    2  2  2 2 2 2 2 9 M1 + 2 2M1 + 1 + 27M1 2M1 M1 + 2 6 − = 0 (14.40) " 54M1 # The author is not aware of any analytical demonstration in the literature which shows that the solution is identical to zero for δ = 016. Nevertheless, this identity can be demonstrated by checking several points for example, M = 1., 2.0, . 1 ∞ Table (14.6) is provided for the following demonstration. Substitution of all the above values into (14.28) results in D = 0. Utilizing the symmetry and antisymmetry of the qualities of the cos and sin for δ < 0 demonstrates that D > 0 regardless of Mach number. Hence, the physical interpretation of this fact is that either no shock exists and the ﬂow is without any discontinuity or that a normal shock exists17. Note that, in the previous case, with a positive large deﬂection angle, there was a transition from one kind of discontinuity to another.

In the range where δ 0, the ≤ X question is whether it is possi- XX XXcoefficientsX ble for an oblique shock to ex- XX a1 a2 a3 M1 XX ist? The answer according to 3 this analysis and stability anal- 1.0 -3 -1 - 2 ysis is no. And according to 9 this analysis, no Mach wave 2.0 3 0 16 can be generated from the wall 1 with zero deflection. In other -1 0 - 16 words, the wall does not emit ∞ any signal to the flow (assuming zero viscosity), which con- Fig. 14.6: The various coefficients of three different tradicts the common approach. Mach numbers to demonstrate that D is zero Nevertheless, in the literature, there are several papers suggesting zero strength Mach wave; others suggest a singular point18. The question of singular point or zero Mach wave strength are only of mathematical interest.

16A mathematical challenge for those who like to work it out. 17There are several papers that attempt to prove this point in the past. Once this analytical solution was published, this proof became trivial. But for non ideal gas (real gas) this solution is only an indication. 18See for example, paper by Rosles, Tabak, “Caustics of weak shock waves,” 206 Phys. Fluids 10 (1) , January 1998. 240 CHAPTER 14. OBLIQUE-SHOCK Suppose that there is a Mach wave at the wall at zero inclination (see Fig- ure (14.7)). Obviously, another Mach wave occurs after a small distance. But because the velocity after a Mach wave (even for an extremely weak shock µ1 µ2 µ3 µ∞ wave) is reduced, thus, the Mach angle will be larger (µ2 > µ1). If the sit- Fig. 14.7: The Mach waves that are supposed uation keeps on occurring over a finite to be generated at zero inclination. distance, there will be a point where the Mach number will be 1 and a normal shock will occur, according the common explanation. However, the reality is that no continuous Mach wave can occur because of the viscosity (boundary layer). In reality, there are imperfections in the wall and in the flow and there is the question of boundary layer. It is well known, in the engineering world, that there is no such thing as a perfect wall. The imperfections of the wall can be, for simplicity’s sake, assumed to be as a sinusoidal shape. For such a wall the zero inclination changes from small positive value to a negative value. If the Mach number is large enough and the wall is rough enough, there will be points where a weak19 weak will be created. On the other hand, the boundary layer covers or smoothens out the bumps. With these conflicting mechanisms, both will not allow a situation of zero inclination with emission of Mach wave. At the very extreme case, only in several points (depending on the bumps) at the leading edge can a very weak shock occur. Therefore, for the purpose of an introductory class, no Mach wave at zero inclination should be assumed. Furthermore, if it was assumed that no boundary layer exists and the wall is perfect, any deviations from the zero inclination angle creates a jump from a positive angle (Mach wave) to a negative angle (expansion wave). This theoretical jump occurs because in a Mach wave the velocity decreases while in the expansion wave the velocity increases. Furthermore, the increase and the decrease depend on the upstream Mach number but in different directions. This jump has to be in reality either smoothened out or has a physical meaning of jump (for example, detach normal shock). The analysis started by looking at a normal shock which occurs when there is a zero inclination. After analysis of the oblique shock, the same conclusion must be reached, i.e. that the normal shock can occur at zero inclination. The analysis of the oblique shock suggests that the inclination angle is not the source (boundary condition) that creates the shock. There must be another boundary condition(s) that causes the normal shock. In the light of this discussion, at least for a simple engineering analysis, the zone in the proximity of zero inclination (small positive and negative inclination angle) should be viewed as a zone without any change unless the boundary conditions cause a normal shock. Nevertheless, emission of Mach wave can occur in other situations. The approxi-

19It is not a mistake, there are two “weaks.” These words mean two different things. The ﬁrst “weak” means more of compression “line” while the other means the weak shock. 14.4. SOLUTION OF MACH ANGLE 241 mation of weak weak wave with nonzero strength has engineering applicability in a very limited cases, especially in acoustic engineering, but for most cases it should be ignored. Oblique Shock

k = 1 4 Mx=3 3 90 0.001

80 2.5 70 0.0005 Myw θs My 2 60 s θw

50 1.5 0 40 0 10 20 30

1 30 -0.0005 20 0.5 10

0 0 -0.001 0.0 10.0 δ 20.0 30.0

Wed Jun 22 15:03:35 2005

Fig. 14.8: The calculation of D (possible error), shock angle, and exit Mach number for M1 = 3

0 10 20 30 δ 14.4.3 Upstream Mach Number, M1, and Shock Angle, θ

The solution for upstream Mach number, M1, and shock angle, θ, are far much simpler and a unique solution exists. The deﬂection angle can be expressed as a function of these variables as

2 (k + 1)M1 cot δ = tan θ 1 (14.41) 2(M 2 sin2 θ 1) − 1 − 242 CHAPTER 14. OBLIQUE-SHOCK or 2 cot θ(M 2 sin2 θ 1) tan δ = 1 − (14.42) 2 + M 2(k + 1 2 sin2 θ) 1 − The pressure ratio can be expressed as P 2kM 2 sin2 θ (k 1) 2 = 1 − − (14.43) P1 k + 1 The density ratio can be expressed as 2 2 ρ2 U1n (k + 1)M1 sin θ = = 2 2 (14.44) ρ1 U2n (k 1)M sin θ + 2 − 1 The temperature ratio expressed as

2 2 2 2 2 T2 c2 2kM1 sin θ (k 1) (k 1)M1 sin θ + 2 = 2 = − − 2 −2 (14.45) T1 c1 (k + 1)M 1 sin θ The Mach number after the shock is 2 2 2 (k 1)M1 sin θ + 2 M2 sin(θ δ) = − (14.46) − 2kM 2 sin2 θ (k 1) 1 − − or explicitly

2 4 2 2 2 2 2 2 (k + 1) M1 sin θ 4(M1 sin θ 1)(kM1 sin θ + 1) M2 = − − (14.47) 2kM 2 sin2 θ (k 1) (k 1)M 2 sin2 θ + 2 1 − − − 1 The ratio of the total pressure can be expressed as

k 1 2 2 k 1 − k 1 P02 (k + 1)M1 sin θ k + 1 − = 2 2 2 2 (14.48) P0 (k 1)M sin θ + 2 2kM sin θ (k 1) 1 − 1 1 − −

Even though the solution for these variables, M1 and θ, is unique, the possible range deﬂection angle, δ, is limited. Examining equation (14.41) shows that the 1 shock angle, θ , has to be in the range of sin− (1/M ) θ (π/2) (see Figure 1 ≥ ≥ 14.9). The range of given θ, upstream Mach number M , is limited between and 1 ∞ 1/ sin2 θ. q 14.4.4 Given Two Angles, δ and θ It is sometimes useful to obtain a relationship where the two angles are known. The ﬁrst upstream Mach number, M1 is 2(cot θ + tan δ) M 2 = (14.49) 1 sin 2θ (tan δ)(k + cos 2θ) − 14.4. SOLUTION OF MACH ANGLE 243

subsonic weak solution

1.0 < M1 < ∞

supersonic strong 1 weak solution −1 θmin = sin soution M1 Defection angle no possible solution solution zone

θ, Shock angle π π θ = 0 θ = θ ∼ 2 max 2

Fig. 14.9: The possible range of solutions for different parameters for given upstream Mach numbers

The reduced pressure difference is

2(P P ) 2 sin θ sin δ 2 − 1 = (14.50) ρU 2 cos(θ δ) − The reduced density is

ρ ρ sin δ 2 − 1 = (14.51) ρ sin θ cos(θ δ) 2 −

For a large upstream Mach number M1 and a small shock angle (yet not approaching zero), θ, the deﬂection angle, δ must also be small as well. Equation (14.41) can be simpliﬁed into

k + 1 θ = δ (14.52) ∼ 2

The results are consistent with the initial assumption which shows that it was an appropriate assumption. 244 CHAPTER 14. OBLIQUE-SHOCK 14.4.5 Flow in a Semi–2D Shape The discussion so far was about the a straight infinite long wedge which is no shock flow direction a “pure” 2–D configuration. Clearly, for any finite length of the wedge, the intermidiaterange analysis { { analysis needs to account for edge ef- normalrange {analysis no shock { fects. The end of the wedge must { have a different configuration (see Fig- ure (14.10)). Yet, the analysis for the middle section produces a close result to reality (because of symmetry). The section where the current analysis is close to reality can be esti- 2-D oblique shock mated from a dimensional analysis for on both sides the required accuracy or by a numer- edge analysis ical method. The dimensional analy- range sis shows that only the doted area to Fig. 14.10: Schematic of finite wedge with zero be area where current solution can be angle of attack. assumed as correctb. In spite of the small area were the current solution can be assumed, this solution is also act as a “reality check” to any numerical analysis. The analysis also provides additional value of the expected range.

aEven finite wedge with limiting wall can be considered as an example for this discussion if the B.L. is neglected. bAt this stage, dimensional analysis is not completed. The author is not aware of any such analysis in literature. The common approach is to carry out numerical analysis. In spite of recent trends, for most engineering applications, a simple tool is sufficient for limit accuracy. Additionally, the numerical works require many times a “reality check.” Another geometry that can be considered as two–dimensional is the cone (some referred to it as Taylor–Maccoll flow). Even though, the cone is a three–dimensional problem, the symmetrical nature of the cone creates a semi–2D problem. In this case there are no edge effects and the geometry dictates slightly different results. The mathematics is much more complicated but there are three solutions. As before, the first solution is thermodynamical unstable. Experimental and analytical work shows that the weak solution is the stable solution and a discussion is provided in the appendix of this chapter. As opposed to the weak shock, the strong shock is unstable, at least, for steady state and no known experiments showing that it exist can be found in the literature. All the literature, known to the author, reports that only a weak shock is possible.

14.4.6 Small δ “Weak Oblique shock”

This interest in this topic is mostly from an academic point of view. It is recommended that this issue be skipped and the time be devoted to other issues. The 14.4. SOLUTION OF MACH ANGLE 245 author is not aware of any single case in which this topic is used in real–world calculations. In fact, after the explicit analytical solution has been provided, studying this topic seems to come at the expense of other more important topics. However, the author admits that as long as there are instructors who examine their students on this issue, it should be covered in this book. For small deﬂection angles, δ, and small normal upstream Mach numbers, M 1 ∼ 1 + ,

1 tan θ = (14.53) M 2 1 1 − p ... under construction.

14.4.7 Close and Far Views of the Oblique Shock In many cases, the close proximity view provides a continuous turning of the deflection angle, δ. Yet, the far view shows a sharp transition. The traditional approach to reconcile these two views is by suggesting that the far view shock is a collection of many small weak shocks (see Figure 14.11). At the local view close to the wall, the oblique shock is a weak “weak oblique” shock. From the far view, the oblique shock is an accumulation of many small (or again weak) “weak shocks.” However, these small “shocks” are built or accumulate into a large and abrupt change (shock). In this theory, the boundary layer (B.L.) does not enter into the cal- θ culation. In reality, the boundary layer increases the zone where a continuous flow exists. The boundary layer reduces δ the upstream flow velocity and therefore the shock does not exist at close proximity to the wall. In larger distance from theFig.wall,14.11:theAshoclocalk becomesand a farpossibviewleof. the oblique shock.

14.4.8 Maximum Value of Oblique shock The maximum values are summarized in the following Table .

Table 14.1: Table of maximum values of the oblique Shock k=1.4

Mx My δmax θmax 1.1000 0.97131 1.5152 76.2762 246 CHAPTER 14. OBLIQUE-SHOCK Table 14.1: Maximum values of oblique shock (continue) k=1.4

Mx My δmax θmax 1.2000 0.95049 3.9442 71.9555 1.3000 0.93629 6.6621 69.3645 1.4000 0.92683 9.4272 67.7023 1.5000 0.92165 12.1127 66.5676 1.6000 0.91941 14.6515 65.7972 1.7000 0.91871 17.0119 65.3066 1.8000 0.91997 19.1833 64.9668 1.9000 0.92224 21.1675 64.7532 2.0000 0.92478 22.9735 64.6465 2.2000 0.93083 26.1028 64.6074 2.4000 0.93747 28.6814 64.6934 2.6000 0.94387 30.8137 64.8443 2.8000 0.94925 32.5875 65.0399 3.0000 0.95435 34.0734 65.2309 3.2000 0.95897 35.3275 65.4144 3.4000 0.96335 36.3934 65.5787 3.6000 0.96630 37.3059 65.7593 3.8000 0.96942 38.0922 65.9087 4.0000 0.97214 38.7739 66.0464 5.0000 0.98183 41.1177 66.5671 6.0000 0.98714 42.4398 66.9020 7.0000 0.99047 43.2546 67.1196 8.0000 0.99337 43.7908 67.2503 9.0000 0.99440 44.1619 67.3673 10.0000 0.99559 44.4290 67.4419

It must be noted that the calculations are for the perfect gas model. In some cases, this assumption might not be sufﬁcient and different analysis is needed. Henderson and Menikoff20 suggested a procedure to calculate the maximum deﬂection angle for arbitrary equation of state21.

14.4.9 Detached shock When the mathematical quantity D becomes positive, for large deﬂection angle, there isn’t a physical solution to an oblique shock. Since the ﬂow “sees” the obstacle, the only possible reaction is by a normal shock which occurs at some distance

20Henderson and Menikoff ”Triple Shock Entropy Theorem” Journal of Fluid Mechanics 366 (1998) pp. 179–210. 21The effect of the equation of state on the maximum and other parameters at this state is unknown at this moment and there are more works underway. 14.4. SOLUTION OF MACH ANGLE 247 from the body. This shock is referred to as the detach shock. The detached shock’s distance from the body is a complex analysis and should be left to graduate class and researchers in this area. Nevertheless, a graph and a general explanation to engineers is provided. Even though this topic has few applications, some might be used in certain situations which the author isn’t aware of. Analysis of the detached shock can be carried out by looking at a body with a round section moving in a su-

personic ﬂow (the absolute velocity weak shock M > 1

9 : ; < = > ? @ ? ACBD< ; E

Strong Shock 8

> @ ;CG isn’t important for this discussion). F Upstream U∞ Figure 14.12 exhibits a round–tip bul- zone B Subsonic Area θ let with a detached shock. The dis- Normal Shock zone A

tance of the detachment is deter- 5 67

mined to a large degree by the up-

H4I J K L M NOI P J'H4Q R J K P L I

T U R V NONWI J U Q L K X Y T Z V R stream Mach number. The zone A S is zone where the flow must be subsonic because at the body the velocity must be zero (the no–slip condition). In such a case, the gas must go through a shock. While at zone Fig. 14.12: The schematic for a round–tip bullet C the flow must be supersonic. The in a supersonic flow weak oblique shock is predicted to flow around the cone. The flow in zone A has to go through some acceleration to became supersonic flow. The explanation to such a phenomenon is above the level of this book (where is the “throat” area question22. Yet, it can be explained as the subsonic is “sucked” into gas in zone C. Regardless of the explanation, these calculations can be summarized by the flowing equation

detachment distance = constant (θ f(M )) (14.54) body thickness × − ∞ where f(M ) is a function of the upstream Mach number which tabulated in the ∞ literature. The constant and the function are different for different geometries. As a general rule, the increase in the upstream Mach results in a decrease of the detachment distance. Larger shock results in a smaller detachment distance, or, alternatively, the ﬂow becomes “blinder” to obstacles. Thus, this phenomenon has a larger impact for a relatively smaller supersonic ﬂow.

14.4.10 Issues Related to the Maximum Deflection Angle The issue of maximum deflection has a practical application aside from the obvious configuration used as a typical simple example. In the typical example, a wedge or

22See example 14.5. 248 CHAPTER 14. OBLIQUE-SHOCK a cone moves into a still medium or gas flows into it. If the deflection angle exceeds the maximum possible, a detached shock occurs. However, there are configurations in which a detached shock occurs in design and engineers need to take it into consideration. Such configurations seem sometimes at first glance not related to the detached shock issue. Consider, for example, a symmetrical suction section in which the deflection angle is just between the maximum deflection angle and above half of the maximum deflection angle. In this situation, at least two oblique shocks occur and after their interaction is shown in Figure (14.13). No detached shock issues are raised when only the first oblique shock is considered. However, the second oblique shock complicates the situation and the second oblique shock can cause a detached shock. This situation is known in the scientific literature as the Mach reflection.

It can be observed that the maxi-

[ \¡] ^ _`¡acbDde[fCghb mum of the oblique shock for the δ perfect gas model depends only on θ1 1 the upstream Mach number i.e., for

U ikj every upstream Mach number there B C θ2 Slip Plane is only one maximum deﬂection an- A gle. δ2

δmax = f(M1) (14.55)

Additionally, it can be observed fFig.or 14.13: The schematic for a symmetrical suc- a maximum oblique shock that a con- tion section with Mach reﬂection

stant deﬂection angle decrease of the

o prqrsutwvrlyx zyt Mach number results in an increase of l/m#n δ Mach angle (weak shock only) M1 > θ1 1 M = θ < θ . The Mach number 2 ⇒ 1 2 decreases after every shock. There- U fore, the maximum deflection angle de- B C creases with a decrease the Mach num- A sub sonic ber. Additionally, due to the symmetry flow a slip plane angle can be guessed to be parallel to original flow, hence δ1 = δ2. Thus, this situation causes the detached shock to appear in the second Fig. 14.14: The “detached” shock in a compli- oblique shock. This detached shock cated configuration sometimes re- manifested itself in a form of curved ferred to as Mach reflection shock (see Figure 14.14). The analysis of this situation is logically very simple, yet the mathematics is somewhat complicated. The maximum deflection angle in this case is, as before, only a function of the upstream Mach number. The calculations for such a case can be carried out by several approaches. It seems that the most straightforward method is the following:

(a) Calculate M1B; 14.4. SOLUTION OF MACH ANGLE 249

(b) Calculate the maximum deﬂection angle, θ2, utilizing (14.31) equation

(d) Use the deﬂection angle, δ2 = δ1 and the Mach number M1B to calculate M1B. Note that no maximum angle is achieved in this shock. POTTO–GDC can be used to calculate this ratio.

This procedure can be extended to calculate the maximum incoming Mach number, M1 by checking the relationship between the intermediate Mach number to M1. In discussing these issues, one must be aware that there are zones of dual solutions in which sharp shock line coexists with a curved line. In general, this zone increases as Mach number increases. For example, at Mach 5 this zone is 8.5◦. For engineering purposes when the Mach number reaches this value, it can be ignored.

14.4.11 Oblique Shock Examples Example 14.2: Air ﬂows at Mach number (M1) or Mx = 4 is approaching a wedge. What is the maximum wedge angle at which the oblique shock can occur? If the wedge angle is 20◦, calculate the weak, the strong Mach numbers, and the respective shock angles.

SOLUTION The maximum wedge angle for (Mx = 4) D has to be equal to zero. The wedge angle that satisﬁes this requirement is by equation (14.28) (a side to the case proximity of δ = 0). The maximum values are:

Mx My δmax θmax 4.0000 0.97234 38.7738 66.0407

To obtain the results of the weak and the strong solutions either utilize the equation (14.28) or the GDC which yields the following results

M M M x ys yw θs θw δ 4.0000 0.48523 2.5686 1.4635 0.56660 0.34907

Example 14.3: A cone shown in Figure (14.15) is exposed to supersonic ﬂow and create an oblique shock. Is the shock shown in the photo weak or strong shock? Explain. Using the geometry provided in the photo, predict at which Mach number was the photo taken based on the assumption that the cone is a wedge. 250 CHAPTER 14. OBLIQUE-SHOCK

Fig. 14.15: Oblique shock occurs around a cone. This photo is courtesy of Dr. Grigory Toker, a Research Professor at Cuernavaco University of Mexico. According to ◦ his measurement, the cone half angle is 15 and the Mach number is 2.2.

SOLUTION The measurement shows that cone angle is 14.43◦ and the shock angle is 30.099◦. With given two angles the solution can be obtained by utilizing equation (14.49) or the Potto-GDC.

P0y M1 My My θs θw δ s w P0x 3.2318 0.56543 2.4522 71.0143 30.0990 14.4300 0.88737

Because the ﬂow is around the cone it must be a weak shock. Even if the cone was a wedge, the shock would be weak because the maximum (transition to a strong shock) occurs at about 60◦. Note that the Mach number is larger than the one predicted by the wedge. 14.4. SOLUTION OF MACH ANGLE 251 Oblique Shock k = 1 4 3 90

80 2.5 70 θ 2 60 δ My 50 1.5 40

1 30

20 0.5 10

0 0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 Mx

Thu Jun 30 15:14:53 2005

Fig. 14.16: Maximum values of the properties in an oblique shock

2 3 4 5 6 7 8 9 10 14.4.12 Application of Oblique ShocMx k

One of the practical applications of the w

oblique shock is the design of an in-

} ~¡ O 4| 4h let suction for a supersonic ﬂow. It is {4|

suggested that a series of weak shocks should replace one normal shock to increase the efficiency (see Figure (14.17))a. Clearly, with a proper design, the flow can be brought to a subsonic Fig. 14.17: Two variations of inlet suction for flow just below M = 1. In such a case, supersonic flow there is less entropy production (less pressure loss). To illustrate the design significance of the oblique shock, the following example is provided.

aIn fact, there is general proof that regardless to the equation of state (any kind of gas), the entropy is to be minimized through a series of oblique shocks rather than through a single normal shock. For details see Henderson and Menikoff “Triple Shock Entropy Theorem,” Journal of Fluid Mechanics 366, (1998) pp. 179–210.

Example 14.4: The Section described in Figure 14.18 air is ﬂowing into a suction section at M = 2.0, P = 1.0[bar], and T = 17◦C. Compare the different conditions in the two 252 CHAPTER 14. OBLIQUE-SHOCK different conﬁgurations. Assume that only a weak shock occurs.

neglect °²± the detached distance

7◦

Normal shock 2 3 4 1 ³µ´ 7◦

Fig. 14.18: Schematic for Example (14.4)

SOLUTION The ﬁrst conﬁguration is of a normal shock for which the results23 are

Ty ρy Py P0y Mx My Tx ρx Px P0x 2.0000 0.57735 1.6875 2.6667 4.5000 0.72087

In the oblique shock, the ﬁrst angle shown is

P0y Mx My My θs θw δ s w P0x 2.0000 0.58974 1.7498 85.7021 36.2098 7.0000 0.99445 and the additional information by the minimal info in the Potto-GDC is

Py Ty P0y Mx My θw δ w Px Tx P0x 2.0000 1.7498 36.2098 7.0000 1.2485 1.1931 0.99445

In the new region, the new angle is 7◦ + 7◦ with new upstream Mach number of Mx = 1.7498 resulting in

P0y Mx My My θs θw δ s w P0x 1.7498 0.71761 1.2346 76.9831 51.5549 14.0000 0.96524

And the additional information is 23The results in this example are obtained using the graphical interface of POTTO–GDC thus, no input explanation is given. In the past the input ﬁle was given but the graphical interface it is no longer needed. 14.4. SOLUTION OF MACH ANGLE 253

Py Ty P0y Mx My θw δ w Px Tx P0x 1.7498 1.5088 41.8770 7.0000 1.2626 1.1853 0.99549

An oblique shock is not possible and normal shock occurs. In such a case, the results are:

Ty ρy Py P0y Mx My Tx ρx Px P0x 1.2346 0.82141 1.1497 1.4018 1.6116 0.98903

With two weak shock waves and a normal shock the total pressure loss is P P P P 04 = 04 03 02 = 0.98903 0.96524 0.99445 = 0.9496 P01 P03 P02 P01 × × The static pressure ratio for the second case is P P P P 4 = 4 3 2 = 1.6116 1.2626 1.285 = 2.6147 P1 P3 P2 P1 × × The loss in this case is much less than in a direct normal shock. In fact, the loss in the normal shock is above than 31% of the total pressure.

Example 14.5: Myw

A∗ A supersonic ﬂow is approaching a very long two– 10◦ dimensional bland wedge body and creates a detached Mys shock at Mach 3.5 (see Figure 14.19). The half wedge angle is 10◦. What is the requited “throat” area ratio to achieve acceleration from the subsonic region Fig. 14.19: Schematic for Ex- to the supersonic region assuming the ﬂow is one– ample (14.5) dimensional?

SOLUTION The detached shock is a normal shock and the results are

Ty ρy Py P0y Mx My Tx ρx Px P0x 3.5000 0.45115 3.3151 4.2609 14.1250 0.21295

Now utilizing the isentropic relationship for k = 1.4 yields

M T ρ A P A P ? × T0 ρ0 A P0 A P0 ∗× 0.45115 0.96089 0.90506 1.4458 0.86966 1.2574

Thus the area ratio has to be 1.4458. Note that the pressure after the weak shock is irrelevant to the area ratio between the normal shock and the “throat” according to the standard nozzle analysis. 254 CHAPTER 14. OBLIQUE-SHOCK Example 14.6: D The effects of a double 4 Slip Plane wedge are explained in the B P3 = P4 3 government web site as weak oblique weak shock oblique shown in Figure (14.20). shock E or expension Adopt this description and wave M assume that the turn of 1 2 0 1 6◦ is made of two equal C angles of 3◦ (see Figure A 14.20). Assume that there are no boundary layers and Fig. 14.20: Schematic of two angles turn with two weak all the shocks are weak and shocks straight. Perform the calculation for M1 = 3.0. Find the required angle of shock BE. Then, explain why this description has internal conﬂict.

SOLUTION The shock BD is an oblique shock with a response to a total turn of 6◦. The conditions for this shock are:

P0y Mx My My θs θw δ s w P0x 3.0000 0.48013 2.7008 87.8807 23.9356 6.0000 0.99105

The transition for shock AB is

P0y Mx My My θs θw δ s w P0x 3.0000 0.47641 2.8482 88.9476 21.5990 3.0000 0.99879

For the shock BC the results are

P0y Mx My My θs θw δ s w P0x 2.8482 0.48610 2.7049 88.8912 22.7080 3.0000 0.99894

And the isentropic relationships for M = 2.7049, 2.7008 are

M T ρ A P A P ? × T0 ρ0 A P0 A P0 ∗× 2.7049 0.40596 0.10500 3.1978 0.04263 0.13632 2.7008 0.40669 0.10548 3.1854 0.04290 0.13665

The combined shocks AB and BC provide the base of calculating the total pressure ratio at zone 3. The total pressure ratio at zone 2 is P P P 02 = 02 01 = 0.99894 0.99879 = 0.997731283 P00 P01 P00 × 14.4. SOLUTION OF MACH ANGLE 255 On the other hand, the pressure at 4 has to be

P P P 4 = 4 04 = 0.04290 0.99105 = 0.042516045 P01 P04 P01 ×

The static pressure at zone 4 and zone 3 have to match according to the government suggestion hence, the angle for BE shock which cause this pressure ratio needs to be found. To do that, check whether the pressure at 2 is above or below or above the pressure (ratio) in zone 4.

P P P 2 = 02 2 = 0.997731283 0.04263 = 0.042436789 P02 P00 P02 ×

Since P2 < P4 a weak shock must occur to increase the static pressure (see P02 P01 Figure 5.4). The increase has to be

P3/P2 = 0.042516045/0.042436789 = 1.001867743

To achieve this kind of pressure ratio the perpendicular component has to be

Ty ρy Py P0y Mx My Tx ρx Px P0x 1.0008 0.99920 1.0005 1.0013 1.0019 1.00000

The shock angle, θ can be calculated from

1 θ = sin− 1.0008/2.7049 = 21.715320879◦

The deﬂection angle for such shock angle with Mach number is

P0y Mx My My θs θw δ s w P0x 2.7049 0.49525 2.7037 0.0 21.72 0.026233 1.00000

From the last calculation it is clear that the government proposed schematic of the double wedge is in conflict with the boundary condition. The flow in zone 3 will flow into the wall in about 2.7◦. In reality the flow of double wedge will produce a curved shock surface with several zones. Only when the flow is far away from the double wedge, the flow behaves as only one theoretical angle of 6◦ exist.

Example 14.7: Calculate the ﬂow deﬂection angle and other parameters downstream when the Mach angle is 34◦ and P1 = 3[bar], T1 = 27◦C, and U1 = 1000m/sec. Assume k = 1.4 and R = 287J/KgK 256 CHAPTER 14. OBLIQUE-SHOCK

SOLUTION The Mach angle of 34◦ is below maximum deﬂection which means that it is a weak shock. Yet, the Upstream Mach number, M1, has to be determined

U1 1000 M1 = = = 2.88 √kRT 1.4 287 300 × × Using this Mach number and the Mach deﬂection in either using the Table or the ﬁgure or POTTO-GDC results in

P0y Mx My My θs θw δ s w P0x 2.8800 0.48269 2.1280 0.0 34.00 15.78 0.89127

The relationship for the temperature and pressure can be obtained by using equation (14.15) and (14.13) or simply converting the M1 to perpendicular component.

M = M sin θ = 2.88 sin(34.0) = 1.61 1n 1 ∗ From the Table (5.1) or GDC the following can be obtained.

Ty ρy Py P0y Mx My Tx ρx Px P0x 1.6100 0.66545 1.3949 2.0485 2.8575 0.89145

The temperature ratio combined upstream temperature yield

T = 1.3949 300 418.5K 2 × ∼ and the same for the pressure

P = 2.8575 3 = 8.57[bar] 2 × And the velocity

U = M √kRT = 2.128√1.4 287 418.5 = 872.6[m/sec] n2 yw × ×

Example 14.8: For Mach number 2.5 and wedge with a total angle of 22◦, calculate the ratio of the stagnation pressure.

Utilizing GDC for Mach number 2.5 and the angle of 11◦ results in

P0y Mx My My θs θw δ s w P0x 2.5000 0.53431 2.0443 85.0995 32.8124 11.0000 0.96873 14.4. SOLUTION OF MACH ANGLE 257 Example 14.9: What is the maximum pressure ratio that can be obtained on wedge when the gas is ﬂowing in 2.5 Mach without any close boundaries? Would it make any difference if the wedge was ﬂowing into the air? If so, what is the difference?

SOLUTION It has to be recognized that without any other boundary condition, the shock is weak shock. For a weak shock the maximum pressure ratio is obtained at the deﬂection point because it is closest to a normal shock. To obtain the maximum point for 2.5 Mach number, either use the Maximum Deﬂection Mach number’s equation or the Potto–GDC

Py Ty P0y Mx My θmax δ max Px Tx P0x 2.5000 0.94021 64.7822 29.7974 4.3573 2.6854 0.60027

In these calculations, Maximum Deﬂection Mach’s equation was used to calculate the normal component of the upstream, then the Mach angle was calculated us- 1 ing the geometrical relationship of θ = sin− M1n/M1. With these two quantities, utilizing equation (14.12) the deﬂection angle, δ, is obtained.

Example 14.10: Consider the schematic shown in the following ﬁgure.

3 stream line 2 1 θ

M1 = 4 δ

Assume that the upstream Mach number is 4 and the deflection angle is δ = 15◦. Compute the pressure ratio and the temperature ratio after the second shock (sometimes referred to as the reflective shock while the first shock is called the incidental shock).

SOLUTION This kind of problem is essentially two wedges placed in a certain geometry. It is clear that the flow must be parallel to the wall. For the first shock, the upstream Mach number is known together with deflection angle. Utilizing the table or the Potto–GDC, the following can be obtained: 258 CHAPTER 14. OBLIQUE-SHOCK

P0y Mx My My θs θw δ s w P0x 4.0000 0.46152 2.9290 85.5851 27.0629 15.0000 0.80382

And the additional information by using minimal information ratio button in Potto– GDC is

Py Ty P0y Mx My θw δ w Px Tx P0x 4.0000 2.9290 27.0629 15.0000 1.7985 1.7344 0.80382

With a Mach number of M = 2.929, the second deﬂection angle is also 15◦. With these values the following can be obtained:

P0y Mx My My θs θw δ s w P0x 2.9290 0.51367 2.2028 84.2808 32.7822 15.0000 0.90041 and the additional information is

Py Ty P0y Mx My θw δ w Px Tx P0x 2.9290 2.2028 32.7822 15.0000 1.6695 1.5764 0.90041

With the combined tables the ratios can be easily calculated. Note that hand calculations requires endless time looking up graphical representation of the solution. Utilizing the POTTO–GDC which provides a solution in just a few clicks. P P P 1 = 1 2 = 1.7985 1.6695 = 3.0026 P3 P2 P3 × T T T 1 = 1 2 = 1.7344 1.5764 = 2.632 T3 T2 T3 ×

Example 14.11: A similar example as before but here Mach angle is 29◦ and Mach number is 2.85. Again calculate the downstream ratios after the second shock and the deﬂection angle.

SOLUTION Here the Mach number and the Mach angle are given. With these pieces of information by utilizing the Potto-GDC the following is obtained:

P0y Mx My My θs θw δ s w P0x 2.8500 0.48469 2.3575 0.0 29.00 10.51 0.96263 14.4. SOLUTION OF MACH ANGLE 259 and the additional information by utilizing the minimal info button in GDC provides

Py Ty P0y Mx My θw δ w Px Tx P0x 2.8500 2.3575 29.0000 10.5131 1.4089 1.3582 0.96263

With the deﬂection angle of δ = 10.51 the so called reﬂective shock gives the following information

P0y Mx My My θs θw δ s w P0x 2.3575 0.54894 1.9419 84.9398 34.0590 10.5100 0.97569 and the additional information of

Py Ty P0y Mx My θw δ w Px Tx P0x 2.3575 1.9419 34.0590 10.5100 1.3984 1.3268 0.97569

P P P 1 = 1 2 = 1.4089 1.3984 1.97 P3 P2 P3 × ∼ T T T 1 = 1 2 = 1.3582 1.3268 1.8021 T3 T2 T3 × ∼

Example 14.12: Compare a direct normal shock to oblique shock with a normal shock. Where will the total pressure loss (entropy) be larger? Assume that upstream Mach number is 5 and the ﬁrst oblique shock has Mach angle of 30◦. What is the deﬂection angle in this case?

SOLUTION For the normal shock the results are

Ty ρy Py P0y Mx My Tx ρx Px P0x 5.0000 0.41523 5.8000 5.0000 29.0000 0.06172

While the results for the oblique shock are

P0y Mx My My θs θw δ s w P0x 5.0000 0.41523 3.0058 0.0 30.00 20.17 0.49901

And the additional information is 260 CHAPTER 14. OBLIQUE-SHOCK

Py Ty P0y Mx My θw δ w Px Tx P0x 5.0000 3.0058 30.0000 20.1736 2.6375 2.5141 0.49901

The normal shock that follows this oblique is

Ty ρy Py P0y Mx My Tx ρx Px P0x 3.0058 0.47485 2.6858 3.8625 10.3740 0.32671

The pressure ratios of the oblique shock with normal shock is the total shock in the second case. P P P 1 = 1 2 = 2.6375 10.374 27.36 P3 P2 P3 × ∼ T T T 1 = 1 2 = 2.5141 2.6858 6.75 T3 T2 T3 × ∼ Note the static pressure raised is less than the combination shocks as compared to the normal shock but the total pressure has the opposite result.

Example 14.13: A ﬂow in a tunnel ends up with two deﬂection angles from both sides (see the following Figure (14.13)).

δ C 2

stream line θ2 D 1 4 slip plane B 0 3 φ

2 θ1 F

stream line A δ1

Illustration for example (14.13)

For upstream Mach number of 5 and deflection angle of 12◦ and 15◦, calculate the pressure at zones 3 and 4 based on the assumption that the slip plane is half of the difference between the two deflection angles. Based on these calculations, explain whether the slip angle is larger or smaller than the difference of the deflection angle.

SOLUTION The ﬁrst two zones immediately after are computed using the same techniques that were developed and discussed earlier. For the ﬁrst direction of 15◦ and Mach number =5. 14.4. SOLUTION OF MACH ANGLE 261

P0y Mx My My θs θw δ s w P0x 5.0000 0.43914 3.5040 86.0739 24.3217 15.0000 0.69317

And the additional conditions are

Py Ty P0y Mx My θw δ w Px Tx P0x 5.0000 3.5040 24.3217 15.0000 1.9791 1.9238 0.69317

For the second direction of 12◦ and Mach number =5.

P0y Mx My My θs θw δ s w P0x 5.0000 0.43016 3.8006 86.9122 21.2845 12.0000 0.80600

And the additional conditions are

Py Ty P0y Mx My θw δ w Px Tx P0x 5.0000 3.8006 21.2845 12.0000 1.6963 1.6625 0.80600

The conditions in zone 4 and zone 3 have two things that are equal. They are the pressure and the velocity direction. It has to be noticed that the velocity magnitudes in zone 3 and 4 do not have to be equal. This non–continuous velocity profile can occur in our model because it is assumed that fluid is non–viscous. If the two sides were equal because of symmetry the slip angle is also zero. It is to say, for the analysis, that only one deflection angle exist. For the two different deflection angles, the slip angle has two extreme cases. The first case is where match lower deflection angle and second is to match the higher deflection angle. In this case, it is assumed that the slip angle moves half of the angle to satisfy both of the deflection angles (first approximation). Under this assumption the conditions in zone 3 are solved by looking at the deflection angle of 12◦ + 1.5◦ = 13.5◦ which results in

P0y Mx My My θs θw δ s w P0x 3.5040 0.47413 2.6986 85.6819 27.6668 13.5000 0.88496 with the additional information

Py Ty P0y Mx My θw δ w Px Tx P0x 3.5040 2.6986 27.6668 13.5000 1.6247 1.5656 0.88496

And in zone 4 the conditions are due to deﬂection angle of 13.5◦ and Mach 3.8006 262 CHAPTER 14. OBLIQUE-SHOCK

P0y Mx My My θs θw δ s w P0x 3.8006 0.46259 2.9035 85.9316 26.3226 13.5000 0.86179

with the additional information

Py Ty P0y Mx My θw δ w Px Tx P0x 3.8006 2.9035 26.3226 13.5000 1.6577 1.6038 0.86179

From these tables the pressure ratio at zone 3 and 4 can be calculated P P P P P 1 1 3 = 3 2 0 1 = 1.6247 1.9791 1.18192 P4 P2 P0 P1 P4 × 1.6963 1.6038 ∼ To reduce the pressure ratio the deflection angle has to be reduced (remember that at weak weak shock almost no pressure change). Thus, the pressure at zone 3 has to be reduced. To reduce the pressure the angle of slip plane has to increase from 1.5 to a larger number. ◦ Example 14.14: The previous example gave rise to another question on the order of the deflection angles. Consider the same values as previous analysis, will the oblique shock with first angle of 15◦ and then 12◦ or opposite order make a difference (M = 5)? If not what order will make a bigger entropy production or pressure loss? (No general proof is needed).

SOLUTION Waiting for the solution

14.4.13 Optimization of Suction Section Design Under heavy construction please ignore The question raised is what is the optimum design for inlet suction unit? There are several considerations that have to be taken into account besides supersonic flow which includes for example the material strength consideration and the operation factors. The optimum deflection angle is a function of the Mach number range in which is presentation of the ex- the suction section is operated in. There are researchers which suggest that the perimental works is useful here? or present the nu- numerical work is the solution. merical works? Perhaps to present the simplified model. 14.5 Summary

As with normal shock, the oblique shock with upstream Mach number, M1 is always greater than 1. However, in oblique, as oppose to the normal shock, the 14.6. APPENDIX: OBLIQUE SHOCK STABILITY ANALYSIS 263 downstream Mach number, M2 could be larger or smaller then 1. The perpendicular component of the downstream Mach number, M1n is always smaller than 1. Given M1 and the deflection angle, δ there could be three solutions: the first one is the “impossible” solution in the case where D is negative, two is weak shock, and three is strong shock. When D is positive there is no physical solution and only normal shock exist. When D is equal to zero, a special case is created because the weak and strong solutions are equal (for large deflection angle). When D > 0, for large deflection angle, there is a possibility of no two–dimensional solution resulting in a detached shock case.

14.6 Appendix: Oblique Shock Stability Analysis

The stability analysis is an analysis which answers the question of what happens if Unstable Stable for some reason, the situation moves away from the expected solution. If the answer turns out to be that Fig. 14.21: Typical examples of unstable and stable situa- the situation will return to its tions original state then it is referred to as the stable situation. On the other hand, if the answer is negative, then the situation is referred to as unstable. An example to this situation, is a ball shown in the Figure (14.21). Instinctively, the stable and unstable can be recognized. There is also the situation where the ball is between the stable and unstable situations when the ball is on a plane field which is referred to as the neutrally stable. In the same manner, the analysis for the oblique shock wave is carried out. The only difference is that here, there are more than one parameter that can be changed, for example, the shock angle, deflection angle, and upstream Mach number. In this example only the weak solution is explained. The similar analysis can be applied to strong shock. Yet, in that analysis it has to be remembered that when the flow becomes subsonic the equation changes from hyperbolic to an elliptic equation. This change complicates the explanation and is omitted in this section. Of course, in the analysis the strong shock results in an elliptic solution (or region) as opposed to a hyperbolic in weak shock. As it results, the discussion is more complicated but similar analysis can be applied to the strong shock. 264 CHAPTER 14. OBLIQUE-SHOCK

The change in the in- ∆θ+ clination angle results ∆θ− in a different upstream Mach number and ∆δ− a different pressure. On the other hand, ∆δ+ to maintain the same direction stream lines, the virtual change in the deﬂection angle Fig. 14.22: The schematic of stability analysis for oblique shock. has to be in the opposite direction of the change of the shock angle. The change is determined from the solution provided before or from the approximation (14.52). k + 1 ∆θ = ∆δ (14.56) 2

+ Equation (14.56) can be applied for either positive, ∆θ or negative ∆θ− values. The pressure difference at the wall becomes a negative increment which tends to pull the shock angle to the opposite direction. The opposite happens when the deflection increment becomes negative, the deflection angle becomes positive which increases the pressure at the wall. Thus, the weak shock is stable. Please note that this analysis doesn’t apply to the case of the close proximity of the δ = 0. In fact, the shock wave is unstable according to this analysis to one direction but stable to the other direction. Yet, it must be pointed out that it doesn’t mean that the flow is unstable but rather that the model is incorrect. There isn’t any known experimental evidence to show that flow is unstable for δ = 0. CHAPTER 15

Prandtl-Meyer Function

positive angle 15.1 Introduction As discussed in Chapter (14) when the deflection turns to the opposite direction of the flow, the flow accelerates to match the maximum angle boundary condition. The transition, as opposed to the oblique shock, is smooth, with- Fig. 15.1: The definition of the angle for out any jump in properties. Here because of the Prandtl–Meyer function. the tradition, the deflection angle is denoted as a positive when it is away from the flow (see Figure (15.1)). In a somewhat a similar concept to oblique shock there exists a “detachment” point above which this model breaks and another model has to be implemented. Yet, when this model breaks down, the flow becomes complicated, flow separation occurs, and no known simple model can describe the situation. As opposed to the oblique shock, there is no limitation for the Prandtl-Meyer function to approach zero. Yet, for very small angles, because of imperfections of the wall and the boundary layer, it has to be assumed to be insignificant.

Supersonic expansion and isentropic com- À U pression (Prandtl-Meyer function), are an exten-

c ¸ sion of the Mach line concept. The Mach line · shows that a disturbance in a field of supersonic flow moves in an angle of µ, which is defined as

(as shown in Figure (15.2)) ¶

º¼»/½¿¾ ¹

1 1 µ = sin− (15.1) M Fig. 15.2: The angles of the Mach line triangle

265 266 CHAPTER 15. PRANDTL-MEYER FUNCTION or

1 1 µ = tan− (15.2) √M 1 1 − A Mach line results because of a small disturbance in the wall contour. This Mach line is assumed to be a result of the positive angle. The reason that a “negative” angle is not applicable is that the coalescing of the small Mach wave which results in a shock wave. However, no shock is created from many small positive angles. The Mach line is the chief line in the analysis because of the wall contour shape information propagates along this line. Once the contour is changed, the flow direction will change to fit the wall. This direction change results in a change of the flow properties, and it is assumed here to be isotropic for a positive angle. This assumption, as it turns out, is close to reality. In this chapter, a discussion on the relationship between the flow properties and the flow direction is presented.

15.2 Geometrical Explanation The change in the ﬂow direction is assume

x ËÊÌ ÍµÎCËÊÌ

to be result of the change in the tangential dx = dUy cos(90 − µ) ÈÊÉ

component. Hence, the total Mach num- Æ Ç

Û Ü4Ý¡ÞÔßÊà á Â ber increases. Therefore, the Mach angle dν

Ã²ÄµÅCÃ dy increase and result in a change in the di- Á

rection of the ﬂow. The velocity compo-

ÐÒÑÔÓCÐ¡ÕOÖ ×4Ø Ï ÓWÙwÕÊÚcÐ nent in the direction of the Mach line is as- Ï sumed to be constant to satisfy the assump- Mach line tion that the change is a result of the contour Fig. 15.3: The schematic of the turning only. Later, this assumption will be exam- flow ined. The typical simplifications for geometrical functions are used: dν sin(dν); (15.3) ∼ cos(dν) 1 ∼ These simplifications are the core reasons why the change occurs only in the perpendicular direction (dν << 1). The change of the velocity in the flow direction, dx is dx = (U + dU) cos ν U = dU (15.4) − In the same manner, the velocity perpendicular to the flow, dy, is dy = (U + dU) sin(dν) = Udν (15.5) The tan µ is the ratio of dy/dx (see Figure (15.3)) dx dU tan µ = = (15.6) dy Udν 15.2. GEOMETRICAL EXPLANATION 267 The ratio dU/U was shown to be

dU dM 2 = (15.7) 2 k 1 2 U 2M 1 + −2 M Combining equations (15.6) and (15.7) tr ansforms itinto

√M 2 1dM 2 dν = − (15.8) 2 k 1 2 −2M 1 + −2 M After integration of equation (15.8) becomes

k + 1 1 k 1 2 1 2 ν(M) = tan− − (M 1) + tan− (M 1) + constant −rk 1 rk + 1 − − − p (15.9)

The constant can be chosen in a such a way that ν = 0 at M = 1.

15.2.1 Alternative Approach to Governing Equations In the previous sec- back tion, a simpliﬁed ver- Mach r Ur line sion was derived based Front Mach on geometrical argu- Uθ line ments. In this section, θ a more rigorous explanation is provided. It must be recognized that here the cylindrical coordinates are advan- tageous because the Fig. 15.4: The schematic of the coordinate based on the mathe- ﬂow turns around a sin- matical description gle point. For this coordinate system, the mass conservation can be written as

∂ (ρrUr) ∂ (ρUθ) + = 0 (15.10) ∂r ∂θ The momentum equations are expressed as

2 2 ∂Ur Uθ ∂Ur Uθ 1 ∂P c ∂ρ U + = = (15.11) r ∂r r ∂θ − r −ρ ∂r − ρ ∂r and ∂U U ∂U U U 1 ∂P c2 ∂ρ U θ + θ θ θ r = = (15.12) r ∂r r ∂θ − r −rρ ∂θ −rρ ∂θ 268 CHAPTER 15. PRANDTL-MEYER FUNCTION If the assumption is that the ﬂow isn’t a function of the radius, r, then all the derivatives with respect to the radius will vanish. One has to remember that when r enters to the function, like the ﬁrst term in the mass equation, the derivative isn’t zero. Hence, the mass equation is reduced to

∂ (ρUθ) ρU + = 0 (15.13) r ∂θ Equation (15.13) can be rearranged as transformed into

1 ∂U 1 ∂ρ U + θ = (15.14) −U r ∂θ ρ ∂θ θ The momentum equations now obtain the form of

U ∂U U 2 θ r θ = 0 r ∂θ − r ∂U U r U = 0 (15.15) θ ∂θ − θ

U ∂U U U c2 ∂ρ θ θ θ r = r ∂θ − r −rρ ∂θ ∂U c2 ∂ρ U θ U = (15.16) θ ∂θ − r − ρ ∂θ 1 ∂ρ Substituting the term ρ ∂θ from equation (15.14) into equation (15.16) results in

2 ∂Uθ c ∂Uθ U U = U + (15.17) θ ∂θ − r U r ∂θ θ or

∂Uθ ∂Uθ U 2 U + = c2 U + (15.18) θ r ∂θ r ∂θ And an additional rearrangement results in

∂Uθ c2 U 2 U + = 0 (15.19) − θ r ∂θ From equation (15.19) it follows that

Uθ = c (15.20)

It is remarkable that the tangential velocity at every turn is at the speed of sound! It must be pointed out that the total velocity isn’t at the speed of sound, but only 15.2. GEOMETRICAL EXPLANATION 269 the tangential component. In fact, based on the definition of the Mach angle, the component shown in Figure (15.3) under Uy is equal to the speed of sound, M = 1. After some additional rearrangement, equation (15.15) becomes U ∂U θ r U = 0 (15.21) r ∂θ − θ If r isn’t approaching infinity, and since U = 0 leads to ∞ θ 6 ∂U r = U (15.22) ∂θ θ In the literature, these results are associated with the characteristic line. This analysis can be also applied to the same equation when they are normalized by Mach number. However, the non–dimensionalization can be applied at this stage as well. The energy equation for any point on a stream line is U 2 + U 2 h(θ) + θ r = h (15.23) 2 0 Enthalpy in perfect gas with a constant specific heat, k, is

c(θ)2

k R 1 C c2 h(θ) = C T = C T = z p}|RT{ = (15.24) p p R (k 1) C k 1 − z}|{v − and substituting this equality, equation (15.24), into equation (15.23) results in c2 U 2 + U 2 + θ r = h (15.25) k 1 2 0 − Utilizing equation (15.20) for the speed of sound and substituting equation (15.22) which is the radial velocity transforms equation (15.25) into

2 2 ∂Ur ∂Ur 2 + Ur ∂θ + ∂θ = h (15.26) k 1 2 0 − After some rearrangement, equation (15.26) becomes

k + 1 ∂U 2 r + U 2 = 2h (15.27) k 1 ∂θ r 0 − Note that Ur must be positive. The solution of the differential equation (15.27) incorporating the constant becomes

k 1 Ur = 2h0 sin θ − (15.28) rk + 1! p 270 CHAPTER 15. PRANDTL-MEYER FUNCTION which satisﬁes equation (15.27) because sin2 θ+cos2 θ = 1. The arbitrary constant in equation (15.28) is chosen such that Ur(θ = 0) = 0. The tangential velocity obtains the form

∂Ur k 1 k 1 Uθ = c = = − 2 h0 cos θ − (15.29) ∂θ rk + 1 rk + 1! p The Mach number in the turning area is

2 2 2 2 2 2 Uθ + Ur Uθ + Ur Ur M = 2 = 2 = 1 + (15.30) c U Uθ θ

Now utilizing the expression that was obtained for Ur and Uθ equations (15.29) and (15.28) results for the Mach number is

k + 1 k 1 M 2 = 1 + tan2 θ − (15.31) k 1 k + 1 − r ! or the reverse function for θ is

k + 1 1 k 1 2 θ = tan− − M 1 (15.32) r k 1 rk + 1 − ! − What happens when the upstream Mach number is not 1? That is when the initial condition for the turning angle doesn’t start with M = 1 but is already at a different angle. The upstream Mach number is denoted in this segment as Mstarting. For this upstream Mach number (see Figure (15.2))

tan ν = M 2 1 (15.33) starting − q The deﬂection angle ν, has to match to the deﬁnition of the angle that is chosen here (θ = 0 when M = 1), so

ν(M) = θ(M) θ(M ) (15.34) − starting

k + 1 1 k 1 2 1 2 = tan− − M 1 tan− M 1 (15.35) rk 1 rk + 1 − ! − − − p p These relationships are plotted in Figure (15.6).

15.2.2 Comparison And Limitations between the Two Ap- proaches The two models produce exactly the same results, but the assumptions for the construction of these models are different. In the geometrical model, the assumption 15.3. THE MAXIMUM TURNING ANGLE 271 is that the velocity change in the radial direction is zero. In the rigorous model, it was assumed that radial velocity is only a function of θ. The statement for the construction of the geometrical model can be improved by assuming that the frame of reference is moving radially in a constant velocity. Regardless of the assumptions that were used in the construction of these models, the fact remains that there is a radial velocity at Ur(r = 0) = constant. At this point (r = 0) these models fail to satisfy the boundary conditions and something else happens there. On top of the complication of the turning point, the question of boundary layer arises. For example, how did the gas accelerate to above the speed of sound when there is no nozzle (where is the nozzle?)? These questions +are of interest in engineering but are beyond the scope of this book (at least at this stage). Normally, the author recommends that this function be used everywhere beyond 2-4 the thickness of the boundary layer based on the upstream length. In fact, analysis of design commonly used in the industry and even questions posted to students show that many assume that the turning point can be sharp. At a small Mach number, (1 + ) the radial velocity is small . However, an increase in the Mach number can result in a very signiﬁcant radial velocity. The radial velocity is “fed” through the reduction of the density. Aside from its close proximity to turning point, mass balance is maintained by the reduction of the density. Thus, some researchers recommend that, in many instances, the sharp point should be replaced by a smoother transition.

15.3 The Maximum Turning Angle The maximum turning angle is obtained when the starting Mach number is 1 and the end Mach number approaches infinity. In this case, Prandtl–Meyer function becomes π k + 1 ν = 1 (15.36) ∞ 2 k 1 − "r − # The maximum of the deflection point and the maximum turning point are only a function of the specific heat ratios. However, the maximum turning angle is much larger than the maximum deflection point because the process is isentropic. What happens when the deflection angel exceeds the maximum angle? The flow in this case behaves as if there is almost a maximum angle and in that region beyond the flow will became vortex street see Figure (15.5)

15.4 The Working Equations for the Prandtl-Meyer Function The change in the deﬂection angle is calculated by ν ν = ν(M ) ν(M ) (15.37) 2 − 1 2 − 1 272 CHAPTER 15. PRANDTL-MEYER FUNCTION

slip line Maximum turning

Fig. 15.5: Expansion of Prandtl-Meyer function when it exceeds the maximum angle

15.5 d’Alembert’s Paradox In ideal inviscid incompressible flows, the move- 3 ment of body does not 1 encounter any resistance. 2 4 This result is known as d’Alembert’s Paradox, and θ w θ2 this paradox is examined 1 θ θ2 here. 1 Supposed that a two– 4 dimensional diamond–shape 1 2 body is stationed in a su- 3 personic flow as shown in Figure (15.7). Again, it is Fig. 15.7: A simplified diamond shape to illustrate the su- assumed that the fluid is in- personic d’Alembert’s Paradox viscid. The net force in flow direction, the drag, is w D = 2 (P P ) = w(P P ) (15.38) 2 2 − 4 2 − 4 It can be observed that only the area that “seems” to be by the flow was used in expressing equation (15.38). The relation between P2 and P4 is such that the flow depends on the upstream Mach number, M1, and the specific heat, k. 15.6. FLAT BODY WITH AN ANGLE OF ATTACK 273 Prandtl-Meyer Angle

100

80 k=1.4

60 θ

0 1 2 3 4 5 6 7 8 9 10 Mach Number

Fri Jul 8 15:39:06 2005

Fig. 15.6: The angle as a function of the Mach number

Regardless in the equation of the state of the gas, the pressure at zone 2, P2, is larger than the pressure at zone 4, P4. Thus, there is always drag when the ﬂow is supersonic which depends on the upstream Mach number, M1, speciﬁc heat, k, and the “visible” area of the object. This drag is known in the literature as (shock) wave drag.

15.6 Flat Body with an Angle of Attack

Previously, the thickness of a body was 1 2 shown to have a drag. Now, a body with 3 zero thickness but with an angle of at- ` tack will be examined. As opposed to w the thickness of the body, in addition 5 4 α to the drag, the body also obtains lift. Slip plane Again, the slip condition is such that the 7 pressure in region 5 and 7 are the same, 6 and additionally the direction of the velocity must be the same. As before, the Fig. 15.8: The definition of the angle for the magnitude of the velocity will be differ- Prandtl–Meyer function. ent between the two regions. 274 CHAPTER 15. PRANDTL-MEYER FUNCTION 15.7 Examples For Prandtl–Meyer Function Example 15.1: A wall is included with 20.0◦ an inclination. A flow of air with a temperature of 20◦C and a speed of U = 450m/sec flows (see Figure 15.9). Calculate the pressure reduction ratio, and the Mach number after the bending point. If the air flows in an imaginary two–dimensional tunnel with width of 0.1[m] what will the width of this imaginary tunnel after the bend? Calculate the “fan” angle. Assume the specific

heat ratio is k = 1.4.

¨ ©

âäãæårçéè/êìë#í¡îyï

ðòñôóéõyöh÷

¥§¦

øúùüûôýyþéÿ ¢¡¤£

Fig. 15.9: The schematic of Example 15.1

SOLUTION First, the initial Mach number has to be calculated (the initial speed of sound). a = √kRT = √1.4 287 293 = 343.1m/sec ∗ ∗ The Mach number is then 450 M = = 1.31 343.1 this Mach number is associated with

M ν P T ρ µ P0 T0 ρ0 1.3100 6.4449 0.35603 0.74448 0.47822 52.6434

The “new” angle should be

ν2 = 6.4449 + 20 = 26.4449◦ and results in

M ν P T ρ µ P0 T0 ρ0 2.0024 26.4449 0.12734 0.55497 0.22944 63.4620 15.7. EXAMPLES FOR PRANDTL–MEYER FUNCTION 275

Note that P01 = P02 P P P 0.12734 2 = 01 2 = = 0.35766 P1 P1 P02 0.35603 The “new” width can be calculated from the mass conservation equation.

ρ M T ρ x M c = ρ x M c = x = x 1 1 1 1 1 1 1 2 2 2 2 2 1 ρ M T ⇒ 2 2 r 2 0.47822 1.31 0.74448 x2 = 0.1 = 0.1579[m] × 0.22944 × 2.0024r0.55497 Note that the compression “fan” stream lines are note and their function can be obtain either by numerical method of going over small angle increments. The other alternative is using the exact solution1. The expansion “fan” angle changes in the Mach angle between the two sides of the bend

fan angle = 63.4 + 20.0 52.6 = 30.8◦ −

Reverse the example, and this time the pressure on both sides are given and the angle has to be obtained2.

Example 15.2:

Gas with k = 1.67 ﬂows over bend (see Figure 15.2). Compute the Mach number

¢!#"$ %

:(;*<>= ?A@ BDC0E>F

6§7 8 9

&('*)+ ,.-0/21 3¢4¤5

Fig. 15.10: The schematic for the reversed question of example (15.2) after the bend, and the bend angle.

1It isn’t really different from this explanation but shown in a more mathematical form, due to Landau and friends. It will be presented in the future version. It isn’t present now because of the low priority to this issue. 2This example is for academic understanding. There is very little with practicality in this kind of problem. 276 CHAPTER 15. PRANDTL-MEYER FUNCTION

SOLUTION The Mach number is determined by satisfying the condition that the pressure downstream are and Mach given. The relative pressure downstream can be calculated by the relationship P P P 1 2 = 2 1 = 0.31424 = 0.2619 P02 P1 P01 1.2 ×

M ν P T ρ µ P0 T0 ρ0 1.4000 7.7720 0.28418 0.60365 0.47077 54.4623

With this pressure ratio P¯ = 0.2619 require either locking in the table or using the enclosed program.

M ν P T ρ µ P0 T0 ρ0 1.4576 9.1719 0.26190 0.58419 0.44831 55.5479

For the rest of the calculation the initial condition is used. The Mach number after the bend is M = 1.4576. It should be noted that speciﬁc heat isn’t k = 1.4 but k = 1.67. The bend angle is

∆ν = 9.1719 7.7720 1.4◦ − ∼

∆µ = 55.5479 54.4623 = 1.0◦ −

15.8 Combination of the Oblique Shock and Isentropic Expansion Example 15.3: Consider two–dimensional flat thin plate at an angle of attack of 4◦ and a Mach number of 3.3. Assume that the specific heat ratio at stage is k = 1.3, calculate the drag coefficient and lift coefficient.

SOLUTION For M = 3.3, the following table can be obtained:

M ν P T ρ µ P0 T0 ρ0 3.3000 62.3113 0.01506 0.37972 0.03965 73.1416

With the angle of attack the region 3 will be at ν 62.31 + 4 for which the following ∼ table can be obtained (Potto-GDC) 15.8. COMBINATION OF THE OBLIQUE SHOCK AND ISENTROPIC EXPANSION277

M ν P T ρ µ P0 T0 ρ0 3.4996 66.3100 0.01090 0.35248 0.03093 74.0528

On the other side, the oblique shock (assuming weak shock) results in

P0y Mx My My θs θw δ s w P0x 3.3000 0.43534 3.1115 88.9313 20.3467 4.0000 0.99676 and the additional information, by clicking on the minimal button, provides

Py Ty P0y Mx My θw δ w Px Tx P0x 3.3000 3.1115 20.3467 4.0000 1.1157 1.1066 0.99676

The pressure ratio at point 3 is P P P P 1 3 = 3 03 01 = 0.0109 1 0.7238 P1 P03 P01 P1 × × 0.01506 ∼ The pressure ratio at point 4 is P 3 = 1.1157 P1

2 2 P4 P3 2 dL = 2 (P4 P3) cos α = 2 cos α = 2 (1.1157 0.7238) cos 4◦ .054 kP M − kM P1 − P1 1.33.3 − ∼ 1 1 1 2 P4 P3 2 dd = 2 sin α = 2 (1.1157 0.7238) sin 4◦ .0039 kM P1 − P1 1.33.3 − ∼ 1 This shows that on the expense of a small drag, a large lift can be obtained. Dis- cussion on the optimum design is left for the next versions. 278 CHAPTER 15. PRANDTL-MEYER FUNCTION CHAPTER 16

Topics in Steady state Two Dimensional ﬂow

shock–expansion theory, linearized potential ﬂow: thin airfoil theory, 2D, method of characteristics

279 280 CHAPTER 16. TOPICS IN STEADY STATE TWO DIMENSIONAL FLOW APPENDIX A

Computer Program

A.1 About the Program The program is written in a C++ language. This program was used to generate all the data in this book. Some parts of the code are in FORTRAN (old code especially for chapters 11 and 12 and not included here.1. The program has the base class of basic ﬂuid mechanics and utilities functions to calculate certain properties given data. The derived class are Fanno, isothermal, shock and others. At this stage only the source code of the program is available no binary available. This program is complied under gnu g++ in /Gnu/Linux system. As much support as possible will be provided if it is in Linux systems. NO Support whatsoever will be provided for any Microsoft system. In fact even PLEASE do not even try to use this program under any Microsoft window system.

A.2 Usage To use the program some information has to be provided. The necessary input parameter(s), the kind of the information needed, where it has to be in a LATEX format or not, and in many case where it is a range of parameter(s). machV The Mach number and it is used in stagnation class

4fL ﬂdV The D and it is used in Fanno class isothermal class p2p1V The pressure ratio of the two sides of the tubes

M1V Entrance Mach M1 to the tube Fanno and isothermal classes

1when will be written in C++ will be add to this program.

281 282 APPENDIX A. COMPUTER PROGRAM real fluids specific functions oblique shock common functions discontinuity normal shock specific functions Rayleigh the actual functions stagnation P-M flow common functions specific functions Isothermal the actual functions CompressibleFlow basic functions virtual functions Interpolation (root finding) LaTeX functions Representation functions

Fig. A.1: Schematicpipe flow diagram that explains the structure of the program common functions Fanno the actual functions only contain A.2. USAGE 283

M1ShockV Entrance Mach M1 when expected shock to the tube Fanno and isothermal classes

FLDShockV FLD with shock in the in Fanno class

4fL M1ﬂdV both M1 and D are given M1ﬂdP2P1V three part info P1 , M and 4fL are given P2 1 D

MxV Mx or My infoStagnation print standard (stagnation) info infoStandard standard info for (Fanno, shock etc) infoTube print tube side info for (Fanno, etc) including infoShock print shock sides info infoTubeShock print tube info shock main info infoTubeProﬁle the Mach number and pressure ratio proﬁles infoTubeShockLimits print tube limits with shock

To get the shock results in LATEX of Mx The following lines have to be inserted in the end of the main function. int isTex = yes; int isRange = no; whatInfo = infoStandard ; variableName = MxV; Mx = 2.0 ; s.makeTable(whatInfo, isRange, isTex, variableName, variableValue);

******************************************* The following stuff is the same as above/below if you use showResults with showHeads but the information is setup for the latex text processing. You can just can cut and paste it in your latex file. You must use longtable style file and dcolumn style files.

******************************************* \setlongtables \begin{longtable} {|D..{1.4}|D..{1.4}|D..{1.4}|D..{1.4}|D..{1.4}|D..{1.4}|D..{1.4}|} \caption{ ?? \label{?:tab:?}}\\ 284 APPENDIX A. COMPUTER PROGRAM \hline \multicolumn{1}{|c|} {$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M} $} & \multicolumn{1}{|c|} {$\mathbf{4fL \over D} $} & \multicolumn{1}{|c|} {$\mathbf{P \over P^{*}} $} & \multicolumn{1}{|c|} {$\mathbf{P_0 \over {P_0}^{*}} $} & \multicolumn{1}{|c|} {$\mathbf{\rho \over \rho^{*}} $} & \multicolumn{1}{|c|} {$\mathbf{U \over {U}^{*}} $} & \multicolumn{1}{|c|} {$\mathbf{T \over T^{*}} $}

\\\hline

\endfirsthead \caption{ ?? (continue)} \\\hline \multicolumn{1}{|c|} {$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M} $} & \multicolumn{1}{|c|} {$\mathbf{4fL \over D} $} & \multicolumn{1}{|c|} {$\mathbf{P \over P^{*}} $} & \multicolumn{1}{|c|} {$\mathbf{P_0 \over {P_0}^{*}} $} & \multicolumn{1}{|c|} {$\mathbf{\rho \over \rho^{*}} $} & \multicolumn{1}{|c|} {$\mathbf{U \over {U}^{*}} $} & \multicolumn{1}{|c|} {$\mathbf{T \over T^{*}} $}

\\\hline \endhead 2.176& 2.152& 0.3608& 1.000& 0.5854& 3.773& 0.6164 \\ \hline\end{longtable}

A.3 Program listings Can be download from www.potto.org. SUBJECTS INDEX 285 Subjects index

A Fliegner experiment, 9 airbag, 200 friction factor, 13 B G Bar-Meir’s solution to Oblique shock, Gibbs, function, 31 8 gravity, 133 Bernoulli’s equation, 28 H C Hydraulic Jump, see discontinuity chamber controled volume, 215 clasifications of chambers, 200 I internal energy, 5 D intersection of Fanno and Rayleigh, 7 Darcy friction factor, 137 isohtermal flow de Laval, Carl Gustaf Patrik, 9 entrace length limitation, 140 defection angle, 226 Isothermal Flow, 2, 3, see Shapiro deflection angle range, 240 flow deLavel’s nozzle, see de Laval, Garl isothermal flow, 135 entrance issues, 140 Gustaf Patrik 4fL diffuser efficiency, 127 maximum , D 139 discontinuity, 1 table, 145 E L Eckert number, 10 large defelection angle, 233 Emanuel’s partial solution to oblique long pipe flow, 135 shock, 8 M External flow, 13 Mach, 3 F maximum deflection angle, 234 Moody diagram, 13 Fanning Friction factor, 137 moving shock, 9 fanno piston velocity, 98 second law, 155 solution for closed valve, 95 Fanno flow, 12 4fL stagnation temperature, 93 fanno flow, 153, D 157 choking, 158 N average friction factor, 159 NACA 1135, 8, 227 entrace Mach number calcula- negative deflection angle, 226 tions, 166 normal components, 228 entropy, 158 nozzle efficiency, 127 shockless, 164 star condition, 160 O Fliegner, 3 oblique shock 286 APPENDIX A. COMPUTER PROGRAM condtions for solution, 231 sub, 114 normal shock, 225 supersonic tunnel, 128 oblique shock governing equations, 229 T Oblique shock stability, 8 table shock choking, 101 P shock wave piston velocity, 98 partial close valve, 107 Taylor–Maccoll flow, 242 R Rayleigh Flow, 12 V rayleigh flow, 187 von Neumann paradox, 225 entrance Mach number, 196 second law, 190 W tables, 191 weak solution, 232 two maximums, 189 Romer, see isothermal nozzle Y Young’s Modulus, 36 S Z science disputes, 5 zero diflection angle, 238 semi rigid chamber, 200 semirigid tank limits, 201 Shapiro Flow, 3 Shapiro flow, 13 shock tube, 109 shock wave, 81 perturbation, 90 solution, 86 star velocity, 87 table basic, 115 thickness, 91 trivail solution, 86 small deflection angles, 242 speed of sound, 4 ideal gas, 29 linear temperature, 31 liquid, 35 real gas, 31 solid, 36 steam table, 30 two phase, 37 speed of sound, what, 28 strong solution, 232 AUTHORS INDEX 287 Authors index

B R Boyle, Robert, 4 Rankine, John Macquorn, 6 Rayleigh, 5 C Riemann, 5 Challis, 5 Rouse, 5 Converdill, 10 S E Shapiro, 4 Eckert, E.R.G, 10 Stodola, 7 Stokes, 5 F Fanno, Gino Girolamo, 7 T Taylor, G. I., 7 G Thompson, 230 Galileo Galilei, 4 V H Van Karman, 4 Henderson, 248 Hugoniot, Pierre Henri, 6 W Wright brothers, 14 K Kutta-Joukowski, 14 L Landau, Lev, 7 Leonardo Da Vinci, 4 M Mach, Ernest, 5 Menikoff, 248 Mersenne, Marin, 4 Meyer, Theodor, 7 Moody, 5 N Newton, 4 O Owczarek, 207 P Poisson, 5, 6 Prandtl, Ludwig, 4, 14